Question

A beaker with 125 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 mol L−1. A student adds 7.00 mLof a 0.360 mol L−1 HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760. Express your answer numerically to two decimal places. Use a minus (−) sign if the pH has decreased.

Answer 1

Acetic acid is written as AcOH and its onjgate base sodium acetate is written as AcO-

Henderson-Hasselbalch equation is

pH = pKa + log

using the given pH and pKa values

5.000 = 4.760 + log

or, log = 5.000 - 4.760

or, log = 0.24

or, = 100.24

or, = 1.74

or, [OAc-] = 1.74 [AcOH]

given,

[OAc-] + [AcOH] = 0.100

or, 1.74 [AcOH] + [AcOH ] = 0.100

or, 2.74 [AcOH ] = 0.100

or , [AcOH ] = (0.100/2.74)

or, [AcOH ] = 0.0365 M

and [OAc-] = (0.100 - 0.0365 ) M = 0.0635 M

now , mmoles of [AcOH ] = molarity volume

= 0.0365 125

= 4.56

and

mmoles of [AcO-] = molarity volume

= 0.0635 125

= 7.94

now mmoles of HCl added

= molarity volume

= 0.360 7.00

= 2.52

Reaction is

AcO- + HCl AcOH + Cl-

BCA table is

mmoles	AcO-	HCl	AcOH
before	7.94	2.52	4.56
change	-2.52	-2.52	+2.52
after	5.42	0	7.08

now, using Henderson-Hasselbalch equation

pH = pKa + log ( mmoles of AcO-/ mmoles of AcOH)

= 4.760 + log(5.42/7.08)

= 4.760 - 0.116

= 4.644

Change in pH

= Final pH - initial pH

= 5.000 - 4.644

= - 0.356

= - 0.36 ( 2 decimal places)