Power of hydrogen ion pH- numerical Part 3

Power of hydrogen ion pH

source : chungcuso3luongyen

Power of hydrogen ion pH-

Question 1) What will be the resultant pH when 200 ml of an aqueous solution of HCl (pH= 2) is mixed with 300 ml of an aqueous solution of NaOH ( pH =12.0 ) ?

Solution )

V of HCl =200 ml

pH of HCl =2

[H⁺] = 10 ^{– pH}= 10 ^-2

Volume of NaOH = 300 ml

pH of  NaOH =  12

pH + pOH = 14

pOH = 14 -12 =2

[OH^–]  = 10 ^-2

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

                                                               = 10 -2  x 200 =  2.0

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

                                                                     = 10 ^-2  x 300 = 3.0

Left Milli equivalents of [OH ^–]  in solution = 3 -2 = 1.0

Total volume = 200 + 300 = 500 ml

[OH^–] = Milli equivalents / Total volume

         = 1 / 500 =2 x 10 ^-3

[OH^–] = 2 x 10 ^-3

[H⁺] [OH^–] = 10 ^-14

[H⁺ ] = 10 ^-14 / 2 x 10 ^{– 3}

[H⁺ ] = 5 x 10 ^-12

pH = – log [H⁺]

      = – log [ 5 x 10^-12]

[log a x b  = log a + log b]

so,

pH  = – [log 5 + log  10^-12]

[log a^b = b log a]

      = – [ 0.6990 – 12 log 10]

because , log 10 = 1

p H  = – [ 0.6990- 12 x 1]

Therefore,

pH  = –  0.6990 +  12 ] = 11.3010

Power of hydrogen ion ‘pH’ = 11.3010  Ans.

Question 2) Calculate Power of hydrogen ion ‘ pH ‘of a solution of given mixture:

a) ( 2 gm CH₃COOH + 3 gm CH₃COONa) in 100 ml of mixture; K_a = 1.8 x 10 ^-5
b) (5 ml of 0.1 M NH₄OH + 250 ml of 0.1 M NH₄Cl) ; K_b = 1.8 x 10 ^-5

c) ( 0.25 mole of acid + 035 mole of salt ) in 500 ml mixture;  K_a =3.6 x 10 ^-4

Solution )

a)

pH = – log K_a + log [salt ] /[acid]

For CH₃COONa,

w = 3 gm , m = 12 + 3 + 12 + 16 x 2 +23= 82

V = 100 ml = 100 /1000 litre =0.1 litre

[CH₃COONa ] = w/mV = 3 / 82 x 0.1 = 0.365

For CH₃COOH,

w = 2 gm , m = 12 + 3 + 12 + 16 x 2 +1 = 60

V = 100 ml = 100 /1000 litre =0.1 litre

[CH₃COOH ] = w/mV = 2 / 60 x 0.1 =  0.333

pH = -log K_a + log [CH₃COONa ] /[CH₃COOH]

p H = – log 1.8 x 10 ^-5 + log [0.365 /0.333]

 pH =- [log 1.8 + log  10^-5] + log  1.096

[log a^b = b log a]

      = – [ 0.2552 – 5 log 10] + 0.0398

because , log 10 = 1

p H  =- [ 0.2552 – 5 x 1] + 0.0398

Therefore,

pH  = –  0.2552 +  5 + 0.0398 =  4.7846

pH = 4.7846 Ans.

b)

pOH = – log K_b + log [salt ] /[base]

pOH = – log K_b + log [NH₄Cl ] / [ NH₄OH]

For NH₄Cl,

N = 0.1, V = 250 ml

milli equivalent of NH₄Cl = 0.1 x 250= 25

For NH4OH,

N = 0.1, V = 5 ml

milli equivalent of NH₄OH = 0.1 x 5= 0.5

[NH₄Cl] = 25 / 255 = 0.098

[NH₄OH] = 0.5 / 255 = 0.002

pOH = – log K_b + log [NH₄Cl ] / [ NH₄OH]

         = – [log 1.8 x 10 ^-5 ] + log [0.098 /0.002]

 pH = – [log 1.8 + log  10^-5] + log  49

[log a^b = b log a]

      = – [ 0.2552 – 5 log 10] + 1.6902

because , log 10 = 1

p H  =- [ 0.2552 – 5 x 1] + 1.6902

Therefore,

pH  = –  0.2552 +  5 + 1.6902 

pH = 6.435 Ans.

c)

pH = – log K_a + log [salt ] /[acid]

V = 500 ml =0.5 litre

[Acid] = no. of mole / volume ( litre)

            = 0.25 / 0.5 =0.5

V = 500 ml =0.5 litre

[salt] = no. of mole / volume ( litre)

           = 0.35 / 0.5 =0.7

pH = – log K_a + log [salt ] /[acid]

p H = – log 3.6 x 10 ^-4 + log [0.7 /0.5]

 pH = – [log 3.6 + log  10^-4] + log  1.4

[log a^b = b log a]

      = – [ 0.5563 – 4 log 10] + 0.1461

because , log 10 = 1

p H  = – [ 0.5563 – 4 x 1] + 0.1461

Therefore,

pH  = –  0.5563 +  4 + 0.1461 =  3.5898

pH = 3.5898 Ans.