Question

Please show work

6. Determine the pH of a 0.35 M aqueous solution of (methylamine). The kb of methylamine is 4.4 x 10-4 1. 10 2. 3.8 3. 12 4. 1.9 _-7. The solubility of manganese(l) hydroxide (Mn(OH)2) is 2.2 x 10-5 M. What is the Ksp of Mn(OH)27 1. 1.1 x 10-14 2, 4.3×10-14 3. 2.1 x 10-14 4, 4.8×10-10 8. For which salt should the aqueous solubility be most sensitive to pH? 1. CaF2 2. Ca(NO3)2 3. CaCI12 4. CaBr2

Answer 1

6) concentration of methyl amine = 0.35M and kb = 4.4x10^-4

Thus pKb of methylamine = -log Kb = 3.356

The pOH of a weak base = 1/2 [pkb - logC]

= 1/2 [3.356 - log 0.35]

= 1.45

Thus pH of the solution = 14 - pOH

= 14-1.45

= 12.55

7) The solubility equilibrium is

Mn(OH)2(s) <-------------> Mn+2(aq) + 2OH- (aq)

- s 2s

Ksp = [Mn+2][OH-]²

= s x (2s)²

= 4s³

Given solubility s = 2.2x10^-5 M Thus

Ksp = 4 (2.2x10^-5 M )³

= 4.259x10^-14

8) The salt whose solubility is most sensitive to pH is CaF2

As CaF2 is a salt of weak acid HF and strong base Ca(OH)2 , itundergoes hydrolysis in aqueous solution and is affected by acid .

The others are salts of strong acids HNO3,HCl and HBr. So their solubility is not affected much by acid.