Adding a Strong Acid to a Buffer?

A beaker with 130mL 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 6.20mL of a 0.290M HCl solution to the beaker. How much will the pH change? The Pk_a of acetic acid is 4.76.Adding a Strong Acid to a Buffer?
5.00 = 4.76 + log [Acetate ] / [acetic acid]

[Acetate ] / [acetic acid] =10^-0.24=0.575

We have to solve the system

Acetate + acid = 0.1

acetate / acid = 0.575



acetate = 0.0362 M

acid = 0.0635 M



Moles acetate = 0.0362 x 130 / 1000 = 0.00471

Moles acid = 0.0635 x 130 / 1000 = 0.00826



Moles HCl added = 6.20 x 0.290 / 1000 = 0.00180

The effect of the added 0.00180 mole H+ ==%26gt; decrease the moles of acetate by 0.00180 and increase the moles of acid by 0.00180 by the reaction H+ + CH3COO- %26gt;%26gt; CH3COOH

Moles acetate =0.00471 - 0.00180 = 0.00291

Moles acid = 0.00826 + 0.00180 = 0.0101

Total volume = 130 + 6.20 = 136.20 mL = 0.1362 L

Concentration acetate = 0.00291 / 0.1362 = 0.0214 M

Concentration acid = 0.0101 / 0.1362 = 0.0742 M



pH = 4.76 + log 0.0214 / 0.0742 = 4.22