How much will the pH change?

A beaker with 190mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M . A student adds 8.70mL of a 0.420M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.How much will the pH change?
We have to determinate the concentration of acetic acid and acetate.

pH = pK + log [CH3COO-] / [CH3COOH]

5.00 = 4.74 + log [CH3COO-] / [CH3COOH]

0.26 = log [CH3COO-] / [CH3COOH]

[CH3COO-] / [CH3COOH ] = 1.82



So we have:

[CH3COO-] / [CH3COOH] = 1.82

[CH3COOH ] + [ CH3COO-] = 0.1



We solve the system and we obtain :

[CH3COOH ] = 0.0355 M

[CH3COO-] = 0.0645 M



Moles CH3COOH in 190 mL = 190 x 0.0355 / 1000 = 0.00675

Moles CH3COO- in 190 mL = 190 x 0.0645 /1000 = 0.0123M

Now we add 8.70 mL HCl 0.420 M

Moles H+ added = 8.70 x 0.420 /1000 = 0.00365

The effect of the added 0.00365 mole of H+ is to decrease the moles of CH3COO- by 0.00365 and increase the moles of CH3COOH by 0.00365 by the reaction

H+ + CH3COO- %26gt;%26gt; CH3COOH

Moles CH3COO- = 0.0123 - 0.00365 = 0.00865

Moles CH3COOH = 0.00675 + 0.00365 = 0.0104

Total volume = 190 + 8.70 = 198.70 mL = 0.1987 L

[ CH3COO-] = 0.00865 / 0.1987 = 0.0435 M

[CH3COOH ] = 0.0104 / 0.1987 = 0.0523 M



pH = 4.74 + log 0.0435/ 0.0523 = 4.66



delta pH = 5.00 - 4.66 = 0.34