solution stoichiometry

Description

Special case

<<<<<<types>>>>>>>

Analysis (stoich)

( )A.P. Chem. Structured Overview Chapter 4

Making

Reacting

like dissolves like

What’s really present (the major species)

Electrolytes (Arrhenius)

non weak strong

Acids and BasesH+ OH-

Quantify (molarity)

Calculations (mol/L)

moles

mass of solute needed

Dilute (M1V1 = M2V2)

PPT

Acid/Base

REDOX

Molec. (formula)

Ionic

Net ionic

gravimetric

volumetrictitration

Solubility rules

oxidation states

process

agents

Balance (both acidic and basic)

(MaVa= MbVb)

(M•V)

9 multiple choice

4 equation writing

3 problem solving

1 short answer

2 one point bonus questions (based on extra little things)

15M (V) = 0.16M (20000.0 mL)

*

Free Response

How much of a 15 M solution would be needed to make 20.0 L of a 0.16 M sol’n?

M1V1 = M2V2

V ≈ 200 mL

* No Calculators on M.C. section; must use mental math

The distinctive odor of vinegar is due to acetic acid, HC2H3O2. Acetic acid reacts with sodium hydroxide in the following fashion:

HC2H3O2(aq) + NaOH(aq) → H2O(l) + NaC2H3O2(aq)

Write the net ionic equation.

If 2.50 mL of vinegar requires 34.9 mL of 0.0960 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00 qt. sample of vinegar?

HC2H3O2(aq) + OH-(aq) → H2O(l) + C2H3O2

-(aq)

MaVa = MbVb

Ma = 0.0960 mol 34.9ml = 1.34 molL L

2.50 mL

( ) ( )

1.34 mol 60.0 g 1 L = L 1 mol 1.06 qts

) ) )( ( ( 75.8 g/qt

Weak acid

Stibnite, (Sb2S3), is the most important ore containing antimony. A sample of the ore was chemically treated to produce antimony (III) ions in solution. The antimony was oxidized to antimony (V) by adding 25.00 mL of 0.0233M KMnO4 solution. The excess KMnO4 was titrated with a 0.0843 M Fe2+ solution; 2.58 mL was required, producing Fe3+ and Mn2+ aqueous ions. All reactions were carried out in acidic solutions. Calculate the mass of the Sb in the sample.

Sb3+Sb5+

MnO4-

Sb5+

Mn2+

Fe3+

Mn2+

Stibnite, (Sb2S3), is the most important ore containing antimony. A sample of the ore was chemically treated to produce antimony (III) ions in solution. The antimony was oxidized to antimony (V) by adding 25.00 mL of 0.0233M KMnO4 solution. The excess KMnO4 was titrated with a 0.0843 M Fe2+ solution; 2.58 mL was required, producing Fe3+ and Mn2+ aqueous ions. All reactions were carried out in acidic solutions. Calculate the mass of the Sb in the sample.

8H+ + MnO4- + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

2.58 mL Fe2+ 0.0843 mol 1 mol MnO4- =

1000 mL 5 mol Fe2+

) ) )( ( ( 0.0000435 mol MnO4

-

(extra)

( () )25.00 mL MnO4- 0.0233 mol =

1000 mL0.000583mol MnO4

-

(total)

0.000583mol (total)

-0.0000435 mol (extra)

0.000540 mol MnO4-(used)

16H+ + 2MnO4- + 5Sb3+ → 2Mn2++ 8H2O + 5Sb5+

( () )0.000540 mol MnO4- 5 mol Sb3+ =

2 mol MnO4-

0.00135 mol Sb3+

(ore)

0.00135 mol Sb3+( )( )121.8 g Sb3+

1 mol Sb3+= 0.164 g Sb3+ (ore)