Question

A beaker with 1.50×10² 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 6.30 mL of a 0.310 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

ANSWER:

According to Henderson-Hesselbaltch equation:

pH = pK_a - log{[HA]/[A^-]}

5.000 = 4.740 - log{[HA]/[A^-]}

log{[HA]/[A^-]} = -0.26

{[HA]/[A^-]} = 0.55

Here,

total molarity of acid and conjugate base in this buffer = [HA] + [A^-] = 0.100 M

and, {[HA]/[A^-]} = 0.55

[HA] = 0.55 x [A^-]

[HA] = 0.55 x {0.100 M - [HA]}

1.55 x [HA] = 0.055 M

[HA] = 0.0355 M

and, [A^-] = {[HA]/(0.55)}

[A^-] = {(0.0355 M)/(0.55)} = 0.0645 M

And, [HA] + [A^-] = 0.0355 M + 0.645 M = 0.100 M

Now,

Number of milimoles of acetic acid = concentration x volume in mL

= 0.0355 M x 1.50 x 10² mL

= 5.325 mmol

Number of milimoles of acetate = concentration x volume in mL

= 0.0645 M x 1.50 x 10² mL

= 9.675 mmol

Number of milimoles of HCl solution added = 0.310 M x 6.30 mL = 1.953 mmol

After addition of HCl solution:

Number of milimoles of acetic acid = 5.325 mmol + 1.953 mmol

= 7.278 mmol

Number of milimoles of acetate = 9.675 mmol - 1.953 mmol

= 7.722 mmol

Total volume = 1.50 x 10² mL + 6.30 mL = 1.563 x 10² mL

And,

concentration of acetic acid = (no. of milimoles)/(total volume in mL)

= (7.278 mmol)/(1.563 x 10² mL)

= 0.0466 M

concentration of acetate = (no. of milimoles)/(total volume in mL)

= (7.722 mmol)/(1.563 x 10² mL)

= 0.0494 M

Now,

According to Henderson-Hesselbaltch equation:

pH = pK_a - log{[HA]/[A^-]}

pH = 4.740 - log{[0.0466 M]/[0.0494 M]}

pH = 4.740 - (-0.025)

pH = 4.765

Therefore, pH change = final pH - initial pH = 4.765 - 5.000 = - 0.235

Hence, pH change due to addition of 6.30 mL of a 0.310 M HCl is 0.235.