Question

A beaker with 1.90×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.00 mL of a 0.300 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740

Answer 1

Let

x = [acetate] and

y = [acetic acid]

5.00 = 4.740 + log x/y

log x/y = 0.26

x/y = 10^0.26 = 1.82

x = 1.82 y

Total molarity of acid and conjugate base in this buffer is 0.100 M

x + y = 0.100

1.82 y + y = 0.1

2.82 y = 0.1

y = 0.035

So, x = 1.82 y = 1.82 x 0.035 = 0.0637

Now, volume 1.90 × 102 mL = 193.8 mL = 0.1938 L

So,

moles acetate      = 0.0637 x 0.1938 L = 0.01234

moles acetic acid = 0.035 x 0.1938 L = 0.0068

Since, HCL is added that is H+ is added, so, concentration of acetic acid will increase and and acetate will decrease.

moles HCl = 0.006 x 0.3 = 0.0018

and

CH₃COO^- + H⁺ = CH₃COOH

moles acetate      = 0.01234 - 0.0018 = 0.01054

moles acetic acid = 0.0068 + 0.0018 =0.0086

Total volume = 193.8 mL + 6.0 mL = 199.8 mL = 0.1998 L = 0.2 L

[acetate]      = 0.01054 / 0.2 = 0.0527 M

[acetic acid ]= 0.0086 / 0.2 = 0.043 M

pH = 4.740 + log (0.0527 /0.043)

     = 4.740 + log (1.225)

     = 4.740 + 0.088

     = 4.828