CHM 2046C Module 11 Chapter 17 Part I Polyprotic Acid calculations

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.

K_a1 = 1.2 x 10^-2 K_a2 = 6.2 x 10^-8

(a) Show both ionization reactions,

H₂SO₃(aq) + H₂O (l) ó H₃O¹⁺ (aq) + HSO₃^1-(aq)

HSO₃^1-(aq) + H₂O (l) ó H₃O¹⁺ (aq) + SO₃^2-(aq)

(b)write the equilibrium expressions for both)

[HSO₃^1-] [H₃O⁺¹]

K_a1 = ------------------------

[ H₂SO₃]

[SO₃^2-] [H₃O⁺¹]

K_a2 = -----------------------

[HSO₃^1-]

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

	[ H₂SO₃]	[HSO₃^1-]	[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 K_a1 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)]^½0.012 ± 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

	[HSO₃^1-]	[H₃O⁺¹]	[SO₃^2-]
initial	0.080M	0.080	0
change	-x	+x	+x
equilibrium	0.080 - x	0.080 + x	x

HSO₃^1-(aq) + H₂O (l) ó H₃O¹⁺ (aq) + SO₃^2-(aq)

[SO₃^2-] [H₃O⁺¹]

K_a2 = -----------------------

[HSO₃^1-]

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 K_a2.

[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

[HSO₃^1-] = 6.2 x 10^-8M, the value of the K_a2