1 relationship between ph and poh the sum of the ph and poh of a solution = 14.00 at 25°c can...
TRANSCRIPT
![Page 1: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/1.jpg)
1
Relationship between pH and pOHRelationship between pH and pOH
the sum of the pH and pOH of a solution = 14.00 at 25°C
can use pOH to find pH of a solution
the sum of the pH and pOH of a solution = 14.00 at 25°C
can use pOH to find pH of a solution
![Page 2: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/2.jpg)
2
pKpK
a way of expressing the strength of an acid or base is pK
pKa = -log(Ka), Ka = 10-pKa
pKb = -log(Kb), Kb = 10-pKb
the stronger the acid, the smaller the pKa
larger Ka = smaller pKa
because it is the –log
a way of expressing the strength of an acid or base is pK
pKa = -log(Ka), Ka = 10-pKa
pKb = -log(Kb), Kb = 10-pKb
the stronger the acid, the smaller the pKa
larger Ka = smaller pKa
because it is the –log
![Page 3: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/3.jpg)
3
Finding the pH of a Strong AcidFinding the pH of a Strong Acid
there are two sources of H3O+ in an aqueous solution of a strong acid – the acid and the water
for the strong acid, the contribution of the water to the total [H3O+] is negligible
for a monoprotic strong acid [H3O+] = [HA]
0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00
there are two sources of H3O+ in an aqueous solution of a strong acid – the acid and the water
for the strong acid, the contribution of the water to the total [H3O+] is negligible
for a monoprotic strong acid [H3O+] = [HA]
0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00
![Page 4: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/4.jpg)
4
Finding the pH of a Weak AcidFinding the pH of a Weak Acid
there are also two sources of H3O+ in and aqueous solution of a weak acid – the acid and the water
however, finding the [H3O+] is complicated by the fact that the acid only undergoes partial ionization
calculating the [H3O+] requires solving an ICE problem for the reaction that defines the acidity of the acid (the top equation)
For a weak acid Ka << 1 so we may be able to apply approximations to avoid solving a quadratic
there are also two sources of H3O+ in and aqueous solution of a weak acid – the acid and the water
however, finding the [H3O+] is complicated by the fact that the acid only undergoes partial ionization
calculating the [H3O+] requires solving an ICE problem for the reaction that defines the acidity of the acid (the top equation)
For a weak acid Ka << 1 so we may be able to apply approximations to avoid solving a quadratic
![Page 5: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/5.jpg)
5
[HNO2] [NO2-] [H3O+]
initial
change
equilibrium
Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
since no products initially, Qc = 0, and the reaction is proceeding forward
HNO2 + H2O NO2- + H3O+
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change
equilibrium
![Page 6: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/6.jpg)
6
[HNO2] [NO2-] [H3O+]
initial 0.200 0 0
change
equilibrium
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
+x+x-x
0.200 -x x x
Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C
![Page 7: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/7.jpg)
7
determine the value of Ka from Table 15.5
since Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.200 x x0.200 -x
Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C
![Page 8: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/8.jpg)
8
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.200 x x
check if the approximation is valid by seeing if x < 5% of [HNO2]init
the approximation is valid
x = 9.6 x 10-3
Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C
![Page 9: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/9.jpg)
9
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.200-x x x
x = 9.6 x 10-3
substitute x into the equilibrium concentration definitions and solve
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.190 0.0096 0.0096
Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C
![Page 10: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/10.jpg)
10
Ka for HNO2 = 4.6 x 10-4
substitute [H3O+] into the formula for pH and solve
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.190 0.0096 0.0096
Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C
![Page 11: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/11.jpg)
11
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.190 0.0096 0.0096
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
though not exact, the answer is reasonably close
Find the pH of 0.200 M HNO2(aq) solution @ 25°CFind the pH of 0.200 M HNO2(aq) solution @ 25°C
![Page 12: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/12.jpg)
Tro, Chemistry: A Molecular Approach
12
What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
![Page 13: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/13.jpg)
13
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
HC6H4NO2 + H2O C6H4NO2- + H3O+
[HA] [A-] [H3O+]
initial 0.012 0 ≈ 0
change
equilibrium
What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
![Page 14: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/14.jpg)
14
[HA] [A-] [H3O+]
initial 0.012 0 0
change
equilibrium
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
+x+x-x
0.012 -x x x
What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
HC6H4NO2 + H2O C6H4NO2- + H3O+
![Page 15: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/15.jpg)
15
determine the value of Ka
since Ka is very small, approximate the [HA]eq = [HA]init and solve for x
[HA] [A2-] [H3O+]
initial 0.012 0 ≈ 0
change -x +x +x
equilibrium 0.012 x x0.012 -x
What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
HC6H4NO2 + H2O C6H4NO2- + H3O+
![Page 16: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/16.jpg)
16
Ka for HC6H4NO2 = 1.4 x 10-5
check if the approximation is valid by seeing if x < 5% of [HC6H4NO2]init
the approximation is valid
x = 4.1 x 10-4
[HA] [A2-] [H3O+]
initial 0.012 0 ≈ 0
change -x +x +x
equilibrium 0.012 x x
What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
![Page 17: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/17.jpg)
17
x = 4.1 x 10-4
substitute x into the equilibrium concentration definitions and solve
[HA] [A2-] [H3O+]
initial 0.012 0 ≈ 0
change -x +x +x
equilibrium 0.012-x x x
What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
![Page 18: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/18.jpg)
18
substitute [H3O+] into the formula for pH and solve
[HA] [A2-] [H3O+]
initial 0.012 0 ≈ 0
change -x +x +x
equilibrium 0.012 0.00041 0.00041
What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
![Page 19: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/19.jpg)
19
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
the values match
[HA] [A2-] [H3O+]
initial 0.012 0 ≈ 0
change -x +x +x
equilibrium 0.012 0.00041 0.00041
What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
![Page 20: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/20.jpg)
20
[HClO2] [ClO2-] [H3O+]
initial 0.100 0 ≈ 0
change
equilibrium
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
HClO2 + H2O ClO2- + H3O+
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
![Page 21: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/21.jpg)
21
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
[HClO2] [ClO2-] [H3O+]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100-x x x
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
![Page 22: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/22.jpg)
Tro, Chemistry: A Molecular Approach
22
Wirte an expression for Ka and use the ICE table to obtain an equation for x
since Ka is very small, approximate the [HClO2]eq = [HClO2]init and solve for x
[HClO2] [ClO2-] [H3O+]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100-x x x
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
![Page 23: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/23.jpg)
23
check if the approximation is valid by seeing if x < 5% of [HNO2]init
the approximation is invalid
x = 3.3 x 10-2
[HClO2] [ClO2-] [H3O+]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100-x x x
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
![Page 24: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/24.jpg)
24
if the approximation is invalid, solve for x using the quadratic formula
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
![Page 25: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/25.jpg)
25
x = 0.028
substitute x into the equilibrium concentration definitions and solve
[HClO2] [ClO2-] [H3O+]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100-x x x
[HClO2] [ClO2-] [H3O+]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.072 0.028 0.028
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
![Page 26: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/26.jpg)
26
substitute [H3O+] into the formula for pH and solve
[HClO2] [ClO2-] [H3O+]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.072 0.028 0.028
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
![Page 27: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/27.jpg)
27
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
the answer matches
[HClO2] [ClO2-] [H3O+]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.072 0.028 0.028
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
Find the pH of 0.100 M HClO2(aq) solution @ 25°C given Ka for HClO2 = 1.1 x 10-2
![Page 28: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/28.jpg)
28
What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25?
What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25?
Use the pH to find the equilibrium [H3O+]
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations and [H3O+]equil
HA + H2O A- + H3O+
[HA] [A-] [H3O+]
initial
change
equilibrium
[HA] [A-] [H3O+]
initial 0.100 0 ≈ 0
change
equilibrium 5.6E-05
![Page 29: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/29.jpg)
29
[HA] [A-] [H3O+]
initial 0.100 0 0
change
equilibrium
fill in the rest of the table using the [H3O+] as a guide
if the difference is insignificant, [HA]equil = [HA]initial
substitute into the Ka expression and compute Ka
+5.6E-05+5.6E-05−5.6E-05
0.100 5.6E-05
5.6E-05 5.6E-050.100
What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25?
What is the Ka of a weak acid if a 0.100 M solution has a pH of 4.25?
HA + H2O A- + H3O+
![Page 30: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/30.jpg)
30
Percent IonizationPercent Ionization
another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water – this is called the percent ionization the higher the percent ionization, the stronger the acid
another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water – this is called the percent ionization the higher the percent ionization, the stronger the acid
• since [ionized acid]equil = [H3O+]equil
![Page 31: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/31.jpg)
31
What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?
What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the Initial Concentrations
Define the Change in Concentration in terms of x
Sum the columns to define the Equilibrium Concentrations
HNO2 + H2O NO2- + H3O+
[HNO2] [NO2-] [H3O+]
initial
change
equilibrium
[HNO2] [NO2-] [H3O+]
initial 2.5 0 ≈ 0
change
equilibrium
+x+x-x
2.5 - x x x
![Page 32: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/32.jpg)
32
determine the value of Ka from Table 15.5
since Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x
[HNO2] [NO2-] [H3O+]
initial 2.5 0 ≈ 0
change -x +x +x
equilibrium 2.5-x ≈2.5 x x
What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?
What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?
![Page 33: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/33.jpg)
33
[HNO2] [NO2-] [H3O+]
initial 2.5 0 ≈ 0
change -x +x +x
equilibrium 2.5 0.034 0.0342.5 - x x x
substitute x into the Equilibrium Concentration definitions and solve
x = 3.4 x 10-2
What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?
What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?
HNO2 + H2O NO2- + H3O+
![Page 34: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/34.jpg)
34
[HNO2] [NO2-] [H3O+]
initial 2.5 0 ≈ 0
change -x +x +x
equilibrium 2.5 0.034 0.034
Apply the Definition and Compute the Percent Ionization
since the percent ionization is < 5%, the “x is small” approximation is valid
What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?
What is the percent ionization of a 2.5 M HNO2 solution if Ka for HNO2 = 4.6 x 10-4 ?
HNO2 + H2O NO2- + H3O+
![Page 35: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/35.jpg)
35
Relationship Between [H3O+]equilibrium and [HA]initialRelationship Between [H3O+]equilibrium and [HA]initial
What happens to [H3O+] at equilibrium if we dilute [HA]?
decreasing/increasing the initial concentration of acid results in increased/decreased percent ionization
this means that the increase in H3O+ concentration is slower than the increase in acid concentration
What happens to [H3O+] at equilibrium if we dilute [HA]?
decreasing/increasing the initial concentration of acid results in increased/decreased percent ionization
this means that the increase in H3O+ concentration is slower than the increase in acid concentration
![Page 36: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/36.jpg)
36
Finding the pH of Mixtures of AcidsFinding the pH of Mixtures of Acids
generally, you can ignore the contribution of the weaker acid to the [H3O+]eq
for a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H3O+] to shift the weak acid equilibrium to the left
so far that the weak acid’s added [H3O+] is negligible
for mixtures of weak acids, generally only need to consider the stronger for the same reasons
as long as one is significantly stronger than the other,
and their concentrations are similar
generally, you can ignore the contribution of the weaker acid to the [H3O+]eq
for a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H3O+] to shift the weak acid equilibrium to the left
so far that the weak acid’s added [H3O+] is negligible
for mixtures of weak acids, generally only need to consider the stronger for the same reasons
as long as one is significantly stronger than the other,
and their concentrations are similar
![Page 37: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/37.jpg)
37
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Write the reactions for the acids with water and determine their Kas
If the Kas are sufficiently different, use the strongest acid to construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
HF + H2O F- + H3O+ Ka = 3.5 x 10-4
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change
equilibrium
HClO + H2O ClO- + H3O+ Ka = 2.9 x 10-8
H2O + H2O OH- + H3O+ Kw = 1.0 x 10-14
![Page 38: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/38.jpg)
38
[HF] [F-] [H3O+]
initial 0.150 0 0
change
equilibrium
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
+x+x-x
0.150 -x x x
![Page 39: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/39.jpg)
39
determine the value of Ka for HF
since Ka is very small, approximate the [HF]eq = [HF]init and solve for x
Ka for HF = 3.5 x 10-4
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.150 x x0.150 -x
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
![Page 40: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/40.jpg)
40
Ka for HF = 3.5 x 10-4
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.150 x x
check if the approximation is valid by seeing if x < 5% of [HF]init
the approximation is valid
x = 7.2 x 10-3
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
![Page 41: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/41.jpg)
41
Ka for HF = 3.5 x 10-4
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.150-x x x
x = 7.2 x 10-3
substitute x into the equilibrium concentration definitions and solve
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.143 0.0072 0.0072
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
![Page 42: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/42.jpg)
42
Ka for HF = 3.5 x 10-4
substitute [H3O+] into the formula for pH and solve
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.143 0.0072 0.0072
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
![Page 43: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/43.jpg)
43
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
though not exact, the answer is reasonably close
[HF] [F-] [H3O+]
initial 0.150 0 ≈ 0
change -x +x +x
equilibrium 0.143 0.0072 0.0072
Ka for HF = 3.5 x 10-4
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO2(aq)
![Page 44: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/44.jpg)
44
NaOH Na+ + OH-
Strong Arrhenius BasesStrong Arrhenius Bases
the stronger the base, the more willing it is to accept H+
use water as the standard acid
for strong bases, practically all molecules are dissociated into OH– or accept H’s
multi-OH strong bases completely dissociated
[HO–] = [strong base] x (# OH)
the stronger the base, the more willing it is to accept H+
use water as the standard acid
for strong bases, practically all molecules are dissociated into OH– or accept H’s
multi-OH strong bases completely dissociated
[HO–] = [strong base] x (# OH)
![Page 45: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/45.jpg)
Calculate the pH at 25°C of a 0.0015 M Sr(OH)2 solution and determine if the solution is acidic, basic, or neutral
Calculate the pH at 25°C of a 0.0015 M Sr(OH)2 solution and determine if the solution is acidic, basic, or neutral
pH is unitless. The fact that the pH > 7 means the solution is basic
[Sr(OH)2] = 1.5 x 10-3 M
pH
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
[H3O+][OH-] pH[Sr(OH)2]
[OH-]=2[Sr(OH)2]
[OH-] = 2(0.0015)= 0.0030 M
![Page 46: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/46.jpg)
46
Practice: Calculate the pH of a 0.0010 M Ba(OH)2
solution and determine if it is acidic, basic, or neutral
![Page 47: 1 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution the sum of the pH](https://reader035.vdocuments.mx/reader035/viewer/2022062222/5697bff11a28abf838cbb814/html5/thumbnails/47.jpg)
47
[H3O+] = 1.00 x 10-14
2.0 x 10-3 = 5.0 x 10-12M
pH > 7 therefore basic
Ba(OH)2 = Ba2+ + 2 OH- therefore [OH-] = 2 x 0.0010M = 0.0020M = 2.0 x 10-3 M
pH = -log [H3O+] = -log (5.0 x 10-12)pH = 11.30
Kw = [H3O+][OH-]
Practice: Calculate the pH of a 0.0010 M Ba(OH)2
solution and determine if it is acidic, basic, or neutral