Question

A 71.0 mL sample of 0.0400 M HIO4 is titrated with 0.0800 M RbOH solution. Calculate the pH after the following volumes of base have been added.

(a) 10.3 mL

pH =

(b) 34.4 mL

pH =

(c) 35.5 mL

pH =

(d) 36.2 mL

pH =

(e) 63.9 mL

pH =

Answer 1

As HIO4 is mono basic acid [H+] = 0.04000M

RbOH is amono acidic base [OH-] = 0.0800M

(a) pH After addition of 10.3 mL of base

From relation:

M_acid X V_acid = M_base X V_base

For milliequivalent of acid: 0.040 X 71.0 = 2.84

milliequivalent of base: = 0.0800 X 10.3 = 0.824

Now, Net effect on reaction medium = 2.84 -0.824 = 2.016 (acidic medium)

So, [H+] = 2.016

pH = -log[H+]

pH = -log[2.016]

After log value, pH = 0.304

Similarly, (b) After addition of 34.4 mL of base:

For milliequivalent of acid: 0.040 X 71.0 = 2.84

milliequivalent of base: = 0.0800 X 34.4 = 2.752

Now, Net effect on reaction medium = 2.84 - 2.752 = 0.088 (still acidic medium)

So, [H+] = 0.088

pH = -log[0.088]

pH = 1.05

(c) After addition of 35.5 mL of base:

For milliequivalent of acid: 0.040 X 71.0 = 2.84

milliequivalent of base: = 0.0800 X 35.5 = 2.84

Now, Net effect on reaction medium = 2.84 - 2.84 = 0.00 (neutral medium)

So, [H+] = [OH-] = 0

pH = 7 (neutral)

(d) After addition of 36.2 mL of base:

For milliequivalent of acid: 0.040 X 71.0 = 2.84

milliequivalent of base: = 0.0800 X 36.2 = 2.896

Now, Net effect om reaction medium = 2.896 - 2.84 = 0.056 (basic medium)

So, [OH-] = 0.056 M

pOH = -log[OH]-

pOH = -log[0.056]

pOH = 1.251

from equation pH + pOH = 14

So, pH = 14 - pOH

pH = 14 - 1.251 = 12.7

(e) After addition of 63.9 mL of base:

For milliequivalent of acid: 0.040 X 71.0 = 2.84

milliequivalent of base: = 0.0800 X 63.9 = 5.112

Now, Net effect on reaction medium = 5.112 - 2.84 = 2.272 (basic medium)

So, [OH-] = 2.272 M

pOH = -log[OH-]

pOH = -log[2.272]

pOH = 0.356

from equation pH + pOH = 14

So, pH = 14 - pOH

pH = 14 - 0.356 = 13.64