## CHM 112 Section 1

Homework Assignment #22
##### Due April 3, 2006 by 8:05 a.m.

Practice (not to be turned in): 16.43, 16.47, 16.51, 16.96, 16.103

Calculate the molar solubility of:

1. Silver nitrite in an X molar solution of silver nitrate where X = the sixth and seventh digits of your Student ID Number divided by 1000.

2. Calcium hydroxide in a buffered solution with a pH = 13.00 + Y where Y = the last two digits of your Student ID Number divided by 100. (If your Student ID Number is 100123456, then the pH = 13.56.)

1. Silver nitrite: AgNO2

 AgNO3(aq) Ag+(aq) + NO3–(aq) AgNO2(s) Ag+(aq) + NO2–(aq) Ksp = [Ag+]e[NO2–]e = 6.0×10–4 Initial X 0 Change + x + x Equilibrium X + x x

Since Ksp is fairly large and X is also pretty large (relatively speaking), no approximation should be made (although this is a close call in many cases).

6.0×10–4 = [X + x]x

x2 + Xx – 6.0×10–4 = 0

x = [–X + (X2 + 4×6.0×10–4)½]/2

x is the molar solubility of silver nitrite in the silver nitrate solution.

2. Calcium hydroxide: Ca(OH)2

The pH gives the hydroxide ion concentration in the solution:

pOH = 14.00 – pH (assuming 25 °C) so pOH = 14.00 – 13.Y = 1.00 – Y

Then, [OH ] = 10pOH = 10(1.00 – Y)

 Ca(OH)2(s) Ca2+(aq) + 2 OH–(aq) Ksp = [Ca2+]e[OH–]e2 = 5.5×10–6 Initial 0 10–(1.00 – Y) Change + x + 2x Equilibrium x 10–(1.00 – Y) + 2x

Since this is a buffered solution, 10–(1.00 – Y) + 2x = 10–(1.00 – Y). Then x = 5.5×10–6/ (10–(1.00 – Y))2, where (in this case) x is the molar solubility of calcium hydroxide.