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Power of hydrogen ion pH-Numerical part 4

Power of hydrogen ion

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Power of hydrogen ion ‘pH’ –

Question 1 ) Calculate [H+] and [OH] concentration in a solution of pH =5 ?

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

[OH ]= 10 -9 Ans.

Question 2) A solution contains 0.0365 gm- ion /litre HCl. Calculate  [H+] and pH of the solution ?

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

pH = 3 Ans.

Question 3)  Dissociation constant of a weak acid (HA) is 4.9 x 10-8. For deci normal solution , Calculate –

(a) % ionisation

(b) pH

(c) OH concentration

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

pH = 4.1549 Ans.

c) [OH ] = ?

[H +] [OH ] = 10 -14

[OH ] = 10-14 / [H +]

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

[OH ]=  1.43 x 10 -10 Ans.

Question 4) Calculate the pH of solution obtained by mixing 10 ml of 0.1 M HCl and 40 ml of 0.2 M H2SO4 ?

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.

Question 5 )  Calculate the pH(Power of hydrogen ion )of a solution which contains 100 ml of 0.1 M HCl and 9.9 ml of 1.0 M NaOH ?

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

pH = 3.0414 Ans.

Question 6) Calculate the pH of a solution at 250C which is twice as alkaline as pure water ?

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

pH = 7.3010  Ans.

 

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