I need help on these 2 questions please. Show work and answer thanks!

Question

Question #1

What is the pH of a solution prepared by dissolving 0.064 moles of benzoic acid (C6H₅COOH)  and 0.062 moles of sodium benzoate in 100 mL of water? (k_a for benzoic acid is 6.3 x 10^-5)

 

Question #2

Calculate the pH of the solute in an aqueous solution of 0.508 M NH₃ (aq) if the k_b is 1.8 x 10^-5

Answer 1

1) First write down the equilibrium reaction for the dissociation of benzoic acid in water. This is

 

C6H5COOH   <=>   H+  + C6H5COO-

 

Then calculate the concentration of benzoic acid and the benzoate ion.

 

[C6H5COOH] = 0.064/0.100 = 0.64 M

 

[C6H5COO-] = 0.062/0.100 = 0.62 M

 

To find the pH you can either set up a ICE table or use the Henderson-Hasselbach equation. The Henderson-Hasselbach equation is   pH = pKa + log([C6H5COO-]/[C6H5COOH]) = 4.19 where pKa = -logKa

 

2) The dissociation of NH3 is

 

NH3 <=> NH4+ + OH-   and the corresponding ICE table is

 

     NH3    <=>   OH-  +  NH4+

I: 0.508             0          0

c: -x                 +x        +x

E: 0.508-x         x          x

 

The equilibrium expression is

 

Kb = [OH-][NH4+]/[NH3] = 1.8x10^-5 = x^2/(0.508-x) = x^2/0.508  since NH3 is a weak base

 

x = [OH-]= 3.0239x10^-3, pOH = -log[OH-] = 2.52

pH = 14 - pOH = 11.48