Power of hydrogen ion pH-
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Power of hydrogen ion pH-
Question ) Calculate Power of hydrogen ion ‘pH’ and pOH of the given bases-
a) 0.01M Ca(OH)2 (b) 0.01 N Ca(OH)2 (c) 0.001 M NaOH or 0.001 N NaOH (d) 0.002 N NaOH (e) 0.006 M Mg(OH)2
Solution)
a) 0.01 M Ca(OH)2 or 10-2 Ca(OH)2
1 M = 2 N
[OH–] = 2 x 10-2
[H+] [OH–] = 10-14
[H+] = 10 -14 /[OH-]
[H+] = 10 -14 / 2 x 10 -2
[H+] = 0.5 x 10-12 = 5 x 10 -13
pH = – log [H+]
= – log [ 5 x 10-13]
[log a x b = log a + log b]
so,
pH = – [log 5 + log 10-13]
[log ab = b log a]
= – [ log 5 – 13 log 10]
because , log 10 = 1
pH =- [ 0.6990 – 13 x 1]
Therefore,
pH = – 0.6990 + 13 x 1 ] = 12.3010
pH = 12.3 Ans.
pH +pOH =14
pOH = 14 – pH
pOH = 14 – 12.3010 = 1.6990
pOH =1.6990 Ans.
Alternative method –
[OH–] = 2 x 10-2
pOH = – log [OH–]
pOH= – log [ 2 x 10-2]
[log a x b = log a + log b]
so,
pOH = – [log 2 + log 10-2]
[log ab = b log a]
= – [ log 2 – 2 log 10]
because , log 10 = 1
pOH =- [ 0.3010 – 2x 1]
Therefore,
pOH = – 0.3010 + 2x 1 ] = 1.6990
pOH = 1.6990 Ans.
pH +pOH =14
pH = 14 – pOH
pH = 14 – 1.6990= 12.3010
pH =12.3 Ans.
b) 0.01 N Ca(OH)2
[OH–] = 10-2
[H+] [OH–] = 10-14
[H+] = 10 -14 /[OH–]
[H+] = 10 -14 / 10 -2
[H+] = 10-12
pH = – log [H+]
= – log [ 10-12]
[log ab = b log a]
= – [ – 12 log 10]
because , log 10 = 1
pH =- [ – 12 x 1] = 12
pH = 12.0 Ans.
pH +pOH =14
pOH = 14 – pH
pOH = 14 – 12.0 = 2.0
pOH = 2.0 Ans.
c) 0.001 M NaOH or 0.001 N NaOH
In both cases ,
[OH-] = 0.001 = 10-3
(because NaOH is mono acidic base. So , M = N)
[OH–] = 10-3
[H+] [OH–] = 10-14
[H+] = 10 -14 /[OH-]
[H+] = 10 -14 / 10 -3
[H+] = 10-11
pH = – log [H+]
= – log [ 10-11]
[log ab = b log a]
= – [ – 11 log 10]
because , log 10 = 1
pH =- [ – 11 x 1] = 11
pH = 11.0 Ans.
pH +pOH =14
pOH = 14 – pH
pOH = 14 – 11.0 = 3.0
pOH = 3.0 Ans.
c) 0.002 N NaOH
[OH-] = 0.002 = 2 x 10-3
[OH–] = 2 x 10-3
[OH-] = 2 x 10-3
pOH = – log [OH–]
pOH= – log [ 2 x 10-3]
[log a x b = log a + log b]
so,
pOH = – [log 2 + log 10-3]
[log ab = b log a]
= – [ log 2 – 3 log 10]
because , log 10 = 1
pOH =- [ 0.3010 – 3 x 1]
Therefore,
pOH = – 0.3010 + 3 x 1 ] = 2.6990
pOH = 2.6990 Ans.
pH +pOH =14
pH = 14 – pOH
pH = 14 – 2.6990= 11.3010
pH =11.3 Ans.
e) 0.006 M Mg(OH)2
It is di acidic base , So
[OH–] = 2 x 0.006
= 0.012 = 1.2 x 10-2
[H+] [OH–] = 10-14
[H+] = 10 -14 /[OH-]
[H+] = 10 -14 / 1.2 x 10 -2
[H+] = 0.83 x 10-12
[H+] = 8.3 x 10-13
pH = – log [H+]
pH= – log [ 8.3 x 10-13]
[log a x b = log a + log b]
so,
pH = – [log 8.3 + log 10-13]
[log ab = b log a]
= – [ log 8.3 – 13 log 10]
because , log 10 = 1
pH =- [ 0.9191 – 13 x 1]
Therefore,
pH = – 0.9191 + 13 = 12.08
pH = 12.08 Ans.
pH +pOH =14
pOH = 14 – pH
pOH = 14 – 12.08 = 1.92
