Question

A beaker with 2.00×10^2 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 6.60 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740

Answer 1

we know that

for buffers

pH = pKa + log [salt / acid ]

in this case

pH = pKa + log [ CH3COO- / CH3COOH ]

5 = 4.74 + log [ CH3COO- / CH3COOH ]

[ CH3COO- / CH3COOH ] = 1.82

[CH3COO-] = 1.82 [CH3COOH]

now

given

total molarity = 0.1

so

[CH3COO-] + [CH3COOH] = 0.1

1.82 [CH3COOH] + [CH3COOH] = 0.1

[CH3COOH] = 0.03546

[CH3COO-] = 1.82 x 0.035461 = 0.06454

now

we know that

moles = molarity x volume (L)

so

moles of HCL added = 0.4 x 6.6 x 10-3 = 2.64 x 10-3

moles of CH3COOH = 0.03546 x 200 x 10-3 = 7.092 x 10-3

moles of Ch3COO- = 0.06456 x 200 x 10-3 = 12.912 x 10-3

now

the reaction is

CH3COO- + H+ --> CH3COOH

we can see that

moles of CH3COO- reacted = moles of H+ added = 2.64 x 10-3

moles of CH3COOH formed = moles of H+ added = 2.64 x 10-3

now

finally

moles of CH3COOH = 7.092 x 10-3 + 2.64 x 10-3 = 9.732 x 10-3

moles of CH3COO- = 12.912 x 10-3 - 2.64 x 10-3 = 10.272 x 10-3

now

pH = pKa + log [ CH3COO- / CH3COOH]

pH = 4.74 + log [ 10.272 x 10-3 / 9.732 x 10-3 ]

pH = 4.76345

so

the pH of the solution is 4.8145

pH change = 4.76345 - 5

pH change = - 0.23655