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26-3-2011

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26-3-2011

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Predicting the direction of a rxnPredict the direction of the following reaction:

HNO2 + CN- HCN + NO2-

Solution: You can either compare between the acids or the bases in both sides

Ka (HNO2) = 4.5*10-4 and Ka(HCN) = 4.9*10-10

HNO2 is a stronger acid which means it is a better proton donor and thus rxn will proceed to right.

Also can say CN- is a stronger base than NO2- ,

therefore CN- is a better proton acceptor and rxn will proceed to right.

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Predict whether the equilibrium constant for the following reaction is greater / or less than 1.

CH3COOH + HCOO- CH3COO- + HCOOH

This is the same as asking whether the reaction proceeds in the forward or backward direction.

Ka(CH3COOH) = 1.8*10-5 and Ka(HCOOH) = 1.7*10-4.

Formic acid is a better acid than acetic acid, therefore the rxn will proceed to left and K<1.

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Predict whether the equilibrium constant for the following reaction is greater / or less than 1.

H3O+(aq) is the strongest acid, it is stronger than acetic acid, therefore, rxn proceeds to left. Ka<1

Or can say: Acetate is a stronger base than H2O and thus it is a better proton acceptor, so the equilibrium favors the left side (K<1).

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO−(aq)

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What does Ka measure?

Ka is a constant used by chemists to compare the relative strength of weak acids.

Many chemists use pKa because it is more convenient.

pKa = -log Ka

For example: the pKa of H2O is 15.7 and the pKa of acetic acid is 4.7. Which is the stronger acid?

The lower the pKa, the stronger the acid.

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HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Weak Acids (HA) and Acid Ionization Constants

HA (aq) H+ (aq) + A- (aq)

Ka =[H+][A-]

[HA]

Ka is the acid ionization constant

Kaweak acidstrength

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What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-]

[HF]= 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka =x2

0.50 - x= 7.1 x 10-4

Ka x2

0.50= 7.1 x 10-4

0.50 – x 0.50Ka << 1

x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M

pH = -log [H+] = 1.72

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When can I use the approximation?

0.50 – x 0.50Ka << 1

When x is less than 5% of the value from which it is subtracted.

x = 0.019Less than 5%??

Approximation ok.RE = (0.019/0.50)*100% = 3.8%

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What is the pH of a 0.05 M HF solution (at 250C)?

Ka x2

0.05= 7.1 x 10-4 x = 0.006 M

0.006 M0.05 M

x 100% = 12%More than 5% !!!!Approximation not

ok.

Must solve for x exactly using quadratic equation or method of successive approximation.

RE =

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Hypochlorous acid is a weak acid formed in laundry bleach. What is the [H3O+] of a 0.125 M HClO solution? Ka = 3.5 x 10-8

Solution:

HClO(aq) + H2O(l) H3O+(aq) + ClO-

(aq)

Ka = = 3.5 x 10-8 [H3O+] [ClO-][HClO]

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Concentration (M) HClO H3O+ ClO-

Initial 0.125 0 0Change -x +x +xEquilibrium 0.125 - x x x

Ka = = 3.5 x 10-8 (x)(x)0.125-x

assume 0.125 - x = 0.125

x2 = 0.4375 x 10-8 x = 6.61 x 10-5

RE = {6.61*10-5/0.125) *100% = 0.053%

[H3O+] = x = 6.61*10-5 M

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What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

Ka =x2

0.122 - x= 5.7 x 10-4

Ka x2

0.122= 5.7 x 10-4

0.122 – x 0.122Ka << 1

x2 = 6.95 x 10-5 x = 0.0083 M

More than 5%??

Approximation not ok.

0.0083 M0.122 M

x 100% = 6.8%RE =

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Ka =x2

0.122 - x = 5.7 x 10-4

x2 + 0.00057x – 6.95 x 10-5 = 0

ax2 + bx + c =0-b ± b2 – 4ac

2ax =

x = 0.0081 x = - 0.0081

[H+] = x = 0.0081 M

pH = -log[H+] = 2.09

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percent ionization = Ionized acid concentration at equilibrium

Initial concentration of acidx 100%

For a monoprotic acid HA

Percent ionization = [H+]

[HA]0

x 100%

[HA]0 = initial concentration

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% Dissociated (also called % Ionized) Weak Acids

• % ionization – a useful way of expressing the strength of an acid or base. – 100% ionized a strong acid. – Only partial ionization occurs with weak acids.

HA(aq) + H2O(l) H3O+(aq) + A (aq)

Initial conc CHA 0 0 to Equil. x +x +x

Equil. CHA x +x +x

%100C

xIonization %

HA

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• Calculate the percent ionization of 0.10 M HCOOH. Ka = 1.7*10-4

• Ka =x2/(0.1 – x), assume 0.1>>x

• X= 4.1*10-3, RE ={4.1*10-3/0.10} * 100 = 4.1%• % ionization = (4.1*10-3/0.10) *100 = 4.1%

HCOOHH+HCOO-

Initial0.1000

Change-xX+X+

Equil0.1-xxx

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Calculate the Percent dissociation of a 0.0100M Hydrocyanic acid solution, Ka = 6.20 x 10-10.

HCN(aq) + H2O(l) H3O+(aq)

+ CN- (aq)

HCN H3O+ CN-

Initial 0.0100M 0 0 Ka = Change -x +x +xFinal 0.0100 –x x x Ka = = 6.20 x 10-10

Assume 0.0100-x = 0.0100 Ka = = 6.2 x 10-10

x = 2.49 x 10-6

% dissociation = x 100% = 0.025%

[H3O+][CN-] [HCN]

(x)(x)(0.0100-x)

x2 0.0100

2.49 x 10-6 0.0100

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Finding the Ka of a Weak Acid from the pH of its Solution

Problem: The weak acid hypochlorous acid is formed in bleach solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the value of the Ka of this weak acid.

Calculating [H3O+] : [H3O+] = 10-pH = 10-4.19 = 6.46 x 10-5 M

Concentration(M) HClO(aq) + H2O(l) H3O+(aq) + ClO -

(aq)

Initial 0.12 ---- ------- -------Change -x ---- +x +xEquilibrium 0.12 -x ---- +x +x

Assumptions: [H3O+] = [H3O+]HClO

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x = [H3O+] = [ClO-] = 6.46 x 10-5 M

Ka = = = 3.48 x 10-8

HClO(aq) + H2O(l) H3O+(aq) + ClO -

(aq)

[H3O+] [ClO-]

[HClO]

(6.46 x 10-5 M)(6.46 x 10-5 M)

0.12 M

Checking:

RE = x 100 = 0.0538 %6.46 x 10-5 M

0.12 M

since HClO is a weak acid, we assume 0.12 M - x = 0.12 M