CHM 2046C Module 11 Chapter 17 Sample Exam

Part G: Common Ion Solution Calculations       10 points

A solution is prepared that is 1.50M HCOOH formic acid 2.0 M Sodium formate. The Ka for formic acids is: 1.8 x 10-4

HCOOH (aq)         +     HOH       <=====>      H3O 1+ (aq)    +    HCOO 1- (aq)

HCOONa (s)       +      HOH       à      Na 1+ (aq)    +    HCOO 1- (aq)

1.    What is the pH of the formic acid before adding the Sodium formate?

HCO2H(aq)   +   H2O (l)  ó    H3O+1 (aq)   +   HCO2-1 (aq)

 [HCO2H] [H3O+1] [HCO2-1] Initial 1.50 M 0 0 Change - x x x Equilibrium 1.50 - x x x
 [HCO2-1] [H3O+1] Ka = ----------------------------- [HCO2H]

Assume x in 1.50 – x is so small you can neglect it. Therefore:

1.50 - ~0 = 1.50

 [x] [x]                x2 1.8 x 10-4 = ------------ = -------- [1.50-x]              1.50-x

1.50 (1.8 x 10-4) = x2

(2.7 x 10-4) = x2         x = 1.34 x 10-2

Assumption: 1.50 - .0134 = 1.49 assumption holds

pH = -Log [H+] = -Log [0.0134] = 1.87

2. What is the pH of Sodium formate solution before mixing?

HCOONa (s)       +      HOH       à      Na 1+ (aq)    +    HCOO 1- (aq)

HCOO 1- (aq)  +      HOH       à   HCOOH  + OH-1

Kw           [HCOOH] [OH 1-]

Kh =   ----  =   ----------------------

Ka               [HCOO1-]

 1.0 x 10-14         [HCOOH] [OH 1-] 5.55 x 10-9 = --------------  =   ---------------------- 1.8 x 10-4                   [HCOO1-]

 [HCOOH] [OH-1] [HCOO1-] Initial 0 0 2.0 Change x x - x Equilibrium x x 2.0-x

 [HCOOH] [OH 1-]          [x] [x] 5.55 x 10-9 = --------------------------- = ----------------------- [HCOO1-]                 [2.0 – x]

Assume x in 2.0 – x is so small you can neglect it. Therefore:

2.0 - ~0 = 2.0

2.0 (5.6 x 10-9) = x2

(10.2 x 10-9) = x2         x = 1.0 x 10-4

Assumption: 2.0 - .00010 = 2.0 assumption holds

pOH = - log(1.0 x 10-4 ) = 4.00  pH = 14.00-pH   pH= 14.00-4.00 = 10.00

3.  What is the pH of the solution of formic acid and  the sodium formate if in each case the volumes of the separate solutions are the same as the volume of the mixture with the concentrations that are listed above.

HCOOH (aq)         +     HOH       <=====>      H3O 1+ (aq)    +    HCOO 1- (aq)

 [HCO2H] [H3O+1] [HCO2-1] Initial 1.50 M 0 2.0 Change - x +x +x Equilibrium 1.50 - x x 2.0 + x
 [HCO2-1] [H3O+1] Ka = ----------------------------- [HCO2H]

 [x] [2.0 + x]           2.0 x 1.8 x 10-4 = ----------------- = -------- [1.50-x]                  1.50

Assume x in (1.50 – x) and the x in (2.0 +x) is so small you can neglect x.

Therefore:

1.50  - ~0 = 1.50  and 2.0 + ~0 = 2.0

x = 1.35 x 10-4

check assumption:

1.50 – x = 1.50 – 0.000135 = ~1.50

2.0 + x = 2.0 + 0.000135 = ~2.0

pH = -Log [H+] = -Log [0.000135] = 3.87