Power of hydrogen ion source : ontrack-media.net

## Power of hydrogen ion ‘pH’ –

### Solution)

pH =5

[H +] =  10 – pH =10-5 gm ion /litre Ans.

[H +] [OH ] = 10 -14

[OH ] = 10 -14 / [H+]

[OH ]= 10 -14 / 10 -5 = 10 -9

### Solution)

S = 0.0365 gm /litre

molecular weight of HCl = 1 + 35.5 = 36.5

HCl is mono basic acid . So basicity = 1

E of  HCl = molecular weight / basicity of acid

E =  36.5 / 1 = 36.5

S (conc. in gm / litre ) = NE

N = S / E

N =  0.0365 /36.5

N = 0.001 = 10 -3

10 -3 N HCl

[H +] = 10 -3

pH = – log [H +]

= – log [ 10 -3]

pH  = – [- 3 log 10 ]

because , log 10 = 1

Therefore,

pH  = –  [- 3 x 1 ] = 3

### Solution )

a) For HA,

K = 2 /V

K = 4.9 x 10-8

deci normal solution means N/10 or 0.1 N solution,

V = 1 / N = 1 /0.1 = 10 litre

∝ = KV . UNDERROOT

∝ = 4.9 x 10 -8 x 10

∝ = 7 x 10 -4

% of ∝ = 7 x 10 -4 x 100 =7 x 10 -2 = 0.07 %

#### % of ∝ = 0.07 % Ans.

b)  [H+] = ∝ / V

= 7 x 10 -4 / 10 = 7 x 10 -5

[H+] = 7 x 10 -5

pH = – log [H+]

= – log [ 7 x 10-5]

[log a x b  = log a + log b]

so,

pH  = – [log 7 + log  10 -5]

[log ab = b log a]

= – [ 0.8451 – 5 log 10]

because , log 10 = 1

p H  =- [ 0.8451 – 5 x 1]

Therefore,

pH  = –  0.8451 + 5 x 1 ] = 4.1549

#### c) [OH –] = ?

[H +] [OH ] = 10 -14

[OH ] = 10-14 / [H +]

[OH ]= 10 -14/ 7 x 10 -5 = 0.143  x 10-9

### Solution )

V of HCl = 10 ml

0.1 M HCl = 0.1 N HCl       (because HCl is mono basic acid)

Milli equivalents of [H+] from HCl = NV(ml)

= 0.1 x 10 = 1

V of H2SO4 = 40 ml

0.2 M H2SO4 = 0.2 x 2 N H2SO4 = 0.4 N H2SO4          (because H2SO4 is dibasic acid)

Milli equivalents of [H+] from H2SO4 = NV(ml)

= 0.4 x 40 =16

Total Milli equivalents of [H+]  in solution =1 + 16 = 17

Total volume = 10 + 40 = 50 ml

[H+] = Milli equivalents / Total volume

= 17 / 50 =3.4 x 10 -1

[H+] = 3.4 x 10-1

pH = – log [H+]

= – log [ 3.4 x 10-1]

[log a x b  = log a + log b]

so,

pH  = – [log 3.4 + log  10-1]

[log ab = b log a]

= – [ 0.5315 – 1 log 10]

because , log 10 = 1

p H  =- [ 0.5315 – 1 x 1]

Therefore,

pH  = –  0.5351 +  1 ] = 0.4649

pH = 0.4649 Ans.

### Solution )

V of HCl = 100 ml

0.1 M HCl = 0.1 N HCl       (because HCl is mono basic acid)

Milli equivalents of [H+] from HCl = NV(ml)

= 0.1 x 100 =10

V of  NaOH  = 9.9 ml

1.0 M NaOH = 1.0  N NaOH           (because NaOH is mono acidic base)

Milli equivalents of [OH] from NaOH = NV(ml)

= 1.0 x 9.9 = 9.9

Milli equivalents of HCl left  in solution =10 – 9.9 = 0.1

Total volume = 100 + 9.9 = 109.9 ml

[H+] = Milli equivalents / Total volume

= 0.1 / 109.9 = 0.000909

[H+] = 9.09  x 10 -4

pH = – log [H+]

= – log [ 9.09  x 10-4]

[log a x b  = log a + log b]

so,

pH  = – [log 9.09 + log  10-4]

[log ab = b log a]

= – [ 0.9586 – 4 log 10]

because , log 10 = 1

p H  =- [ 0.9586- 4 x 1]

Therefore,

pH  = –  0.9586 +  4 ] = 3.0414

#### Solution )

At 250 C,

[OH] = 2 x [OH ] of water

In water,  [H + ] = [OH]= 10 -7

[OH ] = 2.0  x 10 -7

pOH = – log [OH ]

= – log [ 2.0 x 10-7]

[log a x b  = log a + log b]

so,

pOH  = – [log 2 + log  10-7]

[log ab = b log a]

= – [ 0.3010 – 7 log 10]

because , log 10 = 1

pOH  =- [ 0.3010 – 7 x 1]

Therefore,

pOH  = –  0.3010 + 7 x 1 ] = 6.6990

pH + pOH =  14

pH = 14 -pOH

= 14 – 6.6990