A beaker with 165 of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop?

A beaker with 165 mL 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 0.1 M. A student adds 4.40 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pK_a of acetic acid is 4.76.A beaker with 165 of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop?
let x = Molarity acid and let y = molarity acetate

x + y = 0.1



5.00 = 4.74 + log y / x



5.00 = 4.74 + log 0.1 - x / x



10^0.26 = 1.82 = 0.1 - x / x



1.82 x = 0.1 - x

2.82 x = 0.1



x = 0.0355 M

y = 0.1 - 0.0355 = 0.0645 M



moles acid = 0.0355 x 0.165 L = 0.00586

moles acetate = 0.0645 x 0.165 = 0.0106



Moles HCl = 0.400 x 0.0044 L = 0.00176



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



moles acid = 0.00586 + 0.00176 = 0.00762

moles acetate = 0.0106 - 0.00176 = 0.00884



total volume = 165 + 4.4 = 169.4 mL = 0.169 L



concentration acid = 0.00762 / 0.169 = 0.0450 M

concentration acetate = 0.00884 / 0.169 = 0.0523 M



pH = 4.74 + log 0.0523 / 0.0450 = 4.81