# Find the pH of 0.01M sulphuric acid?

Problem.10.3. Find the pH of 0.01M sulphuric acid?

Solution: Sulphuric acid is a strong dibasic acid. It ionizes completely and its one mole produces 2 moles of hydrogen ions.

H2SO4(aq)   —————   2H+(aq) + SO42-(aq)

Therefore,   0.01M   sulphuric   acid   will   produce   2   x   0.01M hydrogen ions.

Hence, hydrogen ions concentration is

[H+] = 2×10-2 M

pH = -log( 2 x 10-2) = _(log 2 + log 10-2)

pH = -log 2 – log 10-2 = – log 2 + 21og 10

pH =-0.3 + 2=1.7

NUMERICALS FROM THE EXERCISE:

Calculate the pH and pOH of 0.2 M H2SO4?

Solution;

Sulphuric acid is a strong dibasic acid. It ionizes completely and its one mole produces 2 moles of hydrogen ions as presented in

equation.

H2SO4 (aq)   ———> 2H+ (aq) + S042 (aq)

Therefore, 0.2M sulphuric acid will produce 2 x 0.2M hydrogen

ions.

Hence, hydrogen ions concentration is

[H+] = 2 x 02M = 4 x 10-1 M

pH   = -log ( 4 x 10-1) = _(log4 + log 10-1)

pH   = -log 4 – log10-1

as- log10-1 = 1

pH   = 1 -Iog4 = 1 – 0.6 = 0.4

1. Calculate the pH of 0.1 MKOH?

Solution:

Potassium hydroxide   solution is   a strong base.   It ionizes completely; one mole of KOH gives one mole of OH- ions.

KOH(aq) —————>  K+(aq) OH(aq)

Therefore, 0.1M solution of KOH produces O.lM OH- ions.

[OH-]  =0.l M or l x 10-1 M

pOH    = -log.(l x 10-1) = -(log 1 + log l0-1)

pOH    = -log l – log l0-1

as log l – 0 and –logl0-1 = 1

pOH     =0+1=1

Therefore, pH of KOH = 14- 1 = 13

1. Calculate the pOH of 0.004 M HNOs?

Solution:

HNO3 is a strong acid so it ionizes completely.

HNO3(aq) —————> H+(aq) + N03(aq)

So, its solution contains 0.004M H+ ions.

[H+] = 4 x 10-3 M

By putting the values of H+ ions in the above equation:

pH = -log(4 x 10-3) = -(Iog4 + loglO-3)

pH = -Iog4 –log 10-3

as-loglO-3 = 3-0.6 = 2.4

pH =3 – 0.6 = 2.4

1. Complete the following Table.
 Solution [H+] [OH-] pH pOH (i) 0.15 M HI 0.15M 0.82 13.18 (“) 0.040 M KOH 0.040 M 12.6 1.4 (iii) 0.020 M Ba (OH)2 0.040 M 12.6 1.4 (iv) 0.00030 M HC1O4 0.00030 M 3.52 10.48 (v) 0.55 M NaOH 0.55 M 13.74 0.26 – Xvi) 0.055 M HC1 0.055 M 1.26 12.74 (vii) 0.055 M Ca(OH)2 0.110M 13.04 0.96

(i) HI ionizes completely as follows:

HI(aq)   ——> H+(aq) + I-(aq)

0.15M HI produces 0.15MH+ ions

[H+] = 15x 10-2 M

pH = -log(15 x 10-2) = -(Iogl5 + 16glO-2)

pH = -Iogl5 -loglO-2

as – log lO-2 = 2

pH = 2 – Iogl5 = 2 – 1.18 = 0.82

(ii) KOH being stronger base ionizes completely as:

KOH(aq) ————— K+(aq) + OH-(aq)

0.040 M KOH produces 0.040 M OH- ions.

[OH-] = 4 x10-2 M

pOH = -log (4 x 10-2) = -(Iog4 + loglO-2)

pOH = -Iog4- loglO-2

as – log lO-2 = 2

pOH = 2 – Iog4 = 2 – 0.6 = 1.4

(iii) Ba (OH)2 ionizes as follows:

Ba (OH)2(aq) —> Ba2+(aq) +  2OH-(aq)

1M Ca (OH)2 produces on ionization twice OH- ions. Hence, 0.020 M Ca (OH)2 produces 2 x 0.020 OH- ions.

[OH-1= 2 x 0.020 = 2 x 2 x 10-2 M = 4 x l0-2 M

pOH = -log (4 x 10-2) = -(log4 + log l0-2) .

pOH = -Iog4 – loglO-2

as – log l0-2=2

pOH= 2 – Iog4 = 2 – 0.6 = 1.4

(iv) HC1O4 ionizes as follows:

HClO4(aq) —————>  H+(aq) + Clo4(aq)

0.00030 M HC1O4 produces 0.0003 M [H+] ions

[H+]=3x 10-4 M.

pH = -log(3 x 10-4) = -(Iog3 + loglO-4)

pH = -log3-loglO-4

as – log lO-4 = 4

pH = 4 – Iog3 = 4 – 0.48 = 3.52

(v) NaOH ionizes as follows:

NaOH(aq)  ——— > Na+(aq)+ OH-

0.55 M NaOH produces 0.55 M OH- ions

[OH-]= 55 x 10-2 M

pOH = -Iog55 – loglO-2

as – log lO-2=2

pOH = 2 – Iog55 = 2 – 1.74 = 0.26

(vi) HCI ionizes as follows:

HCl(aq)   —————> H+(aq) + Cl-(aq)

0.055 M HCI produces 0.055 M H+ ions on ionization.

[H+] = 55 x 10-3M

pH = -log(55 x 1C-3) = -(Iog55 + loglO-3)

pH = -Iog55 -loglO-3.

As – loglO-3 = 3

pH = 3 – Iog55 = 3 – 1.74 = 1.26

(vii) Ca (OH) 2 ionizes as follows:

Ca (OH)2(aq)———Ca2+(aq) + 2OH-(aq)

1M Ca (OH) 2 produces on ionization twice OH- ions. Hence, 0.055 M Ca (OH)2 produces 2 x 0.055 OH-ions.

[OH-]= 2 x 0.055 = 2 x 55 x 1Q-3 M = 110 x 1Q-3 M

pOH = -log (110 x 10-3) = -(log 110 + loglO-3)

pOH –log 1l0 – loglO-3

as – log lO-3 = 3

pOH = 3 – log 1l0 = 3 – 2.04 = 0.96