Question

17. A 25.0-mL sample of 0.20 M NH₃ is titrated with 0.20 M HCl. What is the pH of the solution after 15.00 mL of acid have been added to the ammonia solution? (NH₃ has K_b = 1.8 x 10^–5)

18. The K_sp of silver chloride is 1.6 x 10^–10. (a) Calculate the molar solubility. (b) Dissolve 0.01 moles of NaCl to 1.0 L of water; calculate the molar solubility of AgCl in this salt water.

Answer 1

17. )

millimoles of NH3 = 25 x 0.2 = 5

millimoles of HCl = 15 x 0.2 = 3

pKb = -logKb = -log (1.8 x 10^–5) = 4.74

NH3 + HCl -------------------> NH4Cl

5           3                              0           ----------------> initial

2           0                               3   ----------------------> after reaction

NH3 base NH4Cl salt remains

pOH = pKb + log [salt/base]

pOH = 4.74 + log (3/2)

pOH = 4.92

pH + pOH = 14

pH = 9.08

18.

(a) AgCl -------------------------> Ag+ + Cl-

Ksp = [Ag+][Cl-] = (S)^2

1.6 x 10^-10 = S^2

S = 1.26 x 10^-5 M

molar solubility = 1.26 x 10^-5 M

(b)

NaCl = 0.01 / 1.0 = 0.01 M

     Cl - = 0.01 M

Ksp = [Ag+][Cl] = S x 0.01

1.6 x 10^-10 = S x 0.01

S = 1.6 x 10^-8 M

molar solubility = 1.6 x 10^-8 M