Question

1. (1 ) Calculate the pH during the titration of 50.00 mL of 0.250 M HBr with 0.150 M NaOH after 110.0 mL of NaOH have been added. 2. ) The solubility product for Ce(IO3)3 is 3.2 x 10-10. Calculate the solubility of this compound in water.

Answer 1

1)
Given:
M(HBr) = 0.25 M
V(HBr) = 50 mL
M(NaOH) = 0.15 M
V(NaOH) = 110 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.25 M * 50 mL = 12.5 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.15 M * 110 mL = 16.5 mmol


We have:
mol(HBr) = 12.5 mmol
mol(NaOH) = 16.5 mmol
12.5 mmol of both will react

remaining mol of NaOH = 4 mmol
Total volume = 1.6*10^2 mL

[OH-]= mol of base remaining / volume
[OH-] = 4 mmol/1.6*10^2 mL
= 2.5*10^-2 M


use:
pOH = -log [OH-]
= -log (2.5*10^-2)
= 1.6021


use:
PH = 14 - pOH
= 14 - 1.6021
= 12.3979
Answer: 12.40

2)

At equilibrium:
Ce(IO3)3 <---->     Ce3+     +         3 IO3-  

                     s                  3s      

Ksp = [Ce3+][IO3-]^3
3.2*10^-10=(s)*(3s)^3
3.2*10^-10= 27(s)^4
s = 1.855*10^-3 M
Answer: 1.9*10^-3 M