Power of hydrogen ion

source :designerwater.co.za

## 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,

## 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

## 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

## 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

## 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

## 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