CHM 2046C Module 11 Chapter 17 Sample Exam Answers

Part I:   pH Polyprotic acid                                                    10 points

Sulfurous acid, H2SO3, is a weak acid capable of providing two hydrogen ions.

Ka1 = 1.2 x 10-2    Ka2 = 6.2 x 10-8

(a) Show both ionization reactions,

H2SO3(aq)   +   H2O (l)  ó    H3O1+ (aq)   +   HSO31- (aq)

HSO31- (aq)  +   H2O (l)  ó    H3O1+ (aq)   +   SO32- (aq)

(b)write the equilibrium expressions for both)

 [HSO31-] [H3O+1] Ka1 = ------------------------ [ H2SO3]

 [SO32-] [H3O+1] Ka2 = ----------------------- [HSO31-]

(c) What is the pH of a 0.45 M solution of H2SO3

 [ H2SO3] [HSO31-] [H3O+1] initial 0.45 M 0 0 change -x +x +x equilibrium 0.45 - x x x

 [x] [x] 1.2 x 10-2 = ---------- [0.45-x]

Can not make the assumption as Ka1 is too large

0.012 (0.45 – x) = x2

0.0054 – 0.012x = x2

x2 + 0.012x – 0.0054 = 0

use the quadratic formula to solve for x:

Quadratic formula

A quadratic equation with real or complex coefficients has two (not necessarily distinct) solutions, called roots, which may or may not be real, given by the quadratic formula:

$x = \frac{-b \pm \sqrt {b^2-4ac}}{2a},$

where the symbol "±" indicates that both

 $x_+ = \frac{-b + \sqrt {b^2-4ac}}{2a}$ and $\ x_- = \frac{-b - \sqrt {b^2-4ac}}{2a}$

are solutions.

 0.012 ± [(-0.012)2 –(4∙1∙-0.0054)]½ x = ------------------------------------------ 2(1)

Use only the + in ± below as a real solution

 0.012 ± [(.00014) +(.0216)]½         0.012 ± 0.147     0.159 x = ------------------------------------ = ---------------- = ------- = 0.080 2(1)                                          2                 2

pH = -Log [H+] = -Log [0.080] = 1.10

(d) What is the equilibrium concentration of the sulfite ion,  SO3 2- in the 0.45 M solution of H2SO3

 [HSO31-] [H3O+1] [SO32-] initial 0.080M 0.080 0 change -x +x +x equilibrium 0.080 - x 0.080 + x x

HSO31- (aq)  +   H2O (l)  ó    H3O1+ (aq)   +   SO32- (aq)

 [SO32-] [H3O+1] Ka2 = ----------------------- [HSO31-]

From the first stage ionization the hydrogen ion concentration was found to be very large. The second stage ionization will add very little hydrogen ions to the solution because of the very small value of the Ka2.

 [x] [0.080 +x] 6.2 x 10-8 = ----------------------- [0.080 - x]

Assume x in (0.0.080 – x) and (0.080+x) is so small you can neglect it. Therefore:

 [x] [0.080] 6.2 x 10-8 = ---------------- = x [0.080]

The assumptions hold:

0.080 - .00000008 = 0.080

0.080 +.00000008 = 0.080

[HSO31-] = 6.2 x 10-8 M, the value of the Ka2