Question

Using a 0.20 M phosphate buffer with a pH of 6.0, you add 0.70 mL of 0.52 M HCl to 58 mL of the buffer. What is the new pH of the solution?

Using a 0.30 M phosphate buffer with a pH of 7.1, you add 0.75 mL of 0.48 M NaOH to 45 mL of the buffer. What is the new pH of the solution?

Answer 1

a)

You have a buffer solution that is 1.00M in CH3COOH and 1.00M in CH3COONa.
If you add some NaOH you react with the CH3COOH , reducing its concentration and increasing the concentration of CH3COONa

Mol CH3COOH in 20 mL of 1.00M = 20/1000*1 = 0.02mol CH3COOH
Mol CH3COONa in 20mL of 1.00M = 0.02mol CH2COONa
Mol NaOH in 5.0mL of 0.50M = 5/1000*0.5 = 0.0025mol NaOH
The NaOH reacts with the CH3COOH to form 0.0025 mol CH3COONa and 0.0025 mol CH3COOH is removed
You end up with
CH3COOH = 0.02mol - 0.0025 mol = 0.0175mol CH3COOH in 25mL total solution:
Molarity of CH3COOH = 0.0175/25*1000 = 0.70M CH3COOH
CH3COONa = 0.02mol + 0.0025mol = 0.0225 mol CH3COONa in 25mL total solution
Molarity of CH3COONa = 0.0225/25*1000 = 0.9M CH3COONa

pH = pKa + log([A-]/[HA])
pH = 4.75 + log( 0.9/0.7)
pH = 4.75 + log 1.286
pH = 4.75 + 0.11
pH = 4.86

b)

I used 6.34E-8 for k2.
6.5 = 7.2 + log (base)/(acid)
That gives me (base)/(acid) = 0.2 equation 1.
Equation 2 is acid + base = 0.05M
Solve these two equations simultaneously to obtain (acid) = 0.0417M
(base) = 0.00833M.

Then I change these to millimols.
800 x 0.0417 = 33.36 mmoles acid.
800 x 0.0833 = 6.667 mmols base
You want to add NaOH to make it 7.5
.........acid + OH^- ==> base + H2O
I........33.36....0.......6.667
.add.............x..............
C.........-x.....-x.........x
E.......33.36-x...x.........x

Substitute into the HH equation to obtain
7.5 = 7.2 + log base/acid
7.5 = 7.2 + log (6.667+x)/(33.36-x)
Solve for x = mmoles OH^- needed.
Since M = mmols/mL, plug in to find mL OH^- needed.