Power of hydrogen ion
source :designerwater.co.za
Power of hydrogen ion ‘pH’-
Question 1) Calculate pH(Power of hydrogen ion )of
a)10-3 N HNO3 ( b) 10-3 M H2SO4 . (c) 10-3 N H2SO4 (d) 0.01 N HCl (e) 10 -8 N HCl (f) 10-8 N NaOH
Solution )
Strong acid ionises completely at normal dilution.
a) 10-3 HNO3
[H+] = 10-3
pH = – log [H+]
= – log [ 10-3]
pH = – [- 3 log 10 ]
because , log 10 = 1
Therefore,
pH = – [- 3 x 1 ] = 3
pH = 3 Ans.
b) 10-3 M H2SO4
H2SO4 is dibasic acid,
1 M = 2N
[H+] = 2 x 10-3
pH = – log [H+]
= – log [ 2 x 10-3]
[log a x b = log a + log b]
so,
pH = – [log 2 + log 10-3]
[log ab = b log a]
= – [ 0.3010 – 3 log 10]
because , log 10 = 1
p H =- [ 0.3010 – 3 x 1]
Therefore,
pH = – 0.3010 + 3 x 1 ] = 2.6990
pH = 2.6990 Ans.
c)10-3 NH2SO4
Normality of strong acid = [H+]
[H+] = 10-3
pH = – log [H+]
= – log [ 10-3]
pH = – [log 10-3]
= – [ – 3 log 10]
because , log 10 = 1
p H =- [- 3 x 1] = 3
Therefore,
pH (Power of hydrogen ion ) = 3.0 Ans.
d) 0.01 N HCl
[H+] = 0.01 = 10-2
pH = – log [H+]
= – log [ 10-2]
pH = – [- 2 log 10 ]
because , log 10 = 1
Therefore,
pH = – [- 2 x 1 ] = 3
pH = 2 Ans.
e) 10-8 N HCl
[H+] = 10-8
pH = – log [H+]
= – log [ 10-8]
pH = – [- 8 log 10 ]
because , log 10 = 1
Therefore,
pH = – [- 8 x 1 ] = 8
pH (Power of hydrogen ion )= 8
But pH =8 is not possible in case of HCl because it is an acid and pH of acid should be less than 7.
Correct solution –
[H+] [OH–] = 10-14
In case of water,
[H+] = [OH–]
[H+] of water = 10-7 ( not neglected )
So,
Total [H+] present in 10-8 N HCl = 10-8 + 10-7
[H+] = 0.1 x 10-7 + 1 x 10-7
[H+] =1.1 x 10 -7
pH = – log [H+]
= – log [ 1.1 x 10-7]
[log a x b = log a + log b]
so,
pH = – [log 1.1 + log 10-7]
[log ab = b log a]
= – [ 0.04139 – 7 log 10]
because , log 10 = 1
p H =- [ 0.04139 – 7 x 1]
Therefore,
pH = – 0.04139 + 7 x 1 ] = 6.9586
pH = 6.9586 Ans.
f) 10-8 NaOH
[OH–] = 10-8
[H+] [OH–] = 10-14
[H+] = 10 -14 /[OH-]
[H+] = 10 -14 / 10 -8
[H+] = 10 -6
pH = – log [H+]
= – log [ 10-6]
pH = – [- 6 log 10 ]
because , log 10 = 1
Therefore,
pH = – [- 6 x 1 ] = 6
pH = 6
But pH =6 is not possible in case of NaOH because it is a base and pH(Power of hydrogen ion )of base should be more than 7.
Correct solution –
[H+] [OH–] = 10 -14
In case of water,
[H+] = [OH+]
[OH–] of water = 10 -7 ( not neglected )
So,
Total [OH–] present in 10-8 N NaOH = 10-8 + 10-7
[OH–] = 0.1 x 10-7 + 1 x 10-7
[OH–] =1.1 x 10 -7
pOH = – log [OH–]
= – log [ 1.1 x 10-7]
[log a x b = log a + log b]
so,
pOH = – [log 1.1 + log 10-7]
[log ab = b log a]
= – [ 0.04139 – 7 log 10]
because , log 10 = 1
pOH =- [ 0.04139 – 7 x 1]
Therefore,
pOH = – 0.04139 + 7 x 1 ] = 6.9586
pH + pOH = 14
pH = 14 -pOH
= 14 – 6.9586
pH = 7.0414 Ans.
