Question

1. A mass of sodium pentanoate (CH3CH2 CH2 CH2CO2Na · H2O, FW 124.1135) was dissolved in 50.00 mL of water to make the solution pH = 8.8525. How much was dissolved?

A 30.00 mL portion of 0.4435 M Sn(NO3)2 was titrated with 0.5432 M EDTA at pH = 10.
a. What is the titration reaction and equivalence volume?
b. If Vadded = 0.5Ve what is the pSn2+?
c. What is the p[Titrant] at the equivalence volume?
d. What is the pSn2+ when 26.00 mL of titrant has been added?

Answer 1

1. The general reaction is:
CH₃CH₂CH₂CH₂CO₂^- + H₂O ----------> CH₃CH₂CH₂CH₂CO₂H + OH^-

Now, we know that the pH = 8.8525 that means that the pOH is:
pOH = 14-8.8525 = 5.1475
[OH^-] = 10^-5.1475 = 7.12x10^-6 M

This is the concentration of OH- in equilibrium, from here, we can know how many moles were dissolved innitially. The Ka of pentanoic acid is 1.44x10^-5 so the Kb is: 1x10^-14 / 1.44x10^-5 = 6.94x10^-10

Now according to the ICE chart here:
CH₃CH₂CH₂CH₂CO₂^- + H₂O ----------> CH₃CH₂CH₂CH₂CO₂H + OH^-
e. x - 7.12x10^-6   7.12x10^-6 7.12x10^-6

6.94x10^-10 = (7.12x10^-6)² / x - 7.12x10^-6
6.94x10^-10(x - 7.12x10^-6) = 5.0694x10^-11
x - 7.12x10^-6 = 5.0694x10^-11 / 6.94x10^-10
x = 0.0730467 + 7.12x10^-6
x = 0.07305 M

Now, let's calculate the moles and then the mass:
m = 0.07305 mol/L * 0.050 L * 124.1135 g/mol
m = 0.4533 g

Question 2 post it in another question thread.

Hope this helps