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.
H_{2}SO_{4(aq)} ————— 2H^{+}_{(aq) }+ SO_{4}^{2}_{(aq)}
Therefore, 0.01M sulphuric acid will produce 2 x 0.01M hydrogen ions.
Hence, hydrogen ions concentration is
[H+] = 2×102 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 H_{2}SO_{4}?
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.
H_{2}SO_{4}_{ (aq)} ———> 2H^{+} _{(aq) }+ S0_{4}^{2} _{(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
 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
 Calculate the pOH of 0.004 M HNOs?
Solution:
HNO_{3} is a strong acid so it ionizes completely.
HNO_{3(aq)} —————> H^{+}_{(aq)} + N0_{3}–_{(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}
asloglO^{3 }= 30.6 = 2.4
pH =3 – 0.6 = 2.4
 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 HC1O_{4}

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 x102 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.
[OH1= 2 x 0.020 = 2 x 2 x 102 M = 4 x l0^{2} M
pOH = log (4 x 102) = (log4 + log l0^{2}) .
pOH = Iog4 – loglO^{2}
as – log l0^{2}=2
pOH= 2 – Iog4 = 2 – 0.6 = 1.4
(iv) HC1O_{4} ionizes as follows:
HClO_{4(aq) }—————> H^{+}_{(aq) }+ Clo_{4}–_{(aq) }
0.00030 M HC1O_{4} produces 0.0003 M [H^{+}] ions
[H^{+}]=3x 104 M.
pH = log(3 x 10^{4}) = (Iog3 + loglO^{4})
pH = log3loglO^{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 102 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 103M
pH = log(55 x 1C^{3}) = (Iog55 + loglO3)
pH = Iog55 loglO3.
As – loglO^{3} = 3
pH = 3 – Iog55 = 3 – 1.74 = 1.26
(vii) Ca (OH)_{ 2} ionizes as follows:
Ca (OH)_{2}(aq)———Ca^{2+}_{(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 OHions.
[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