Power of hydrogen ion pH source : chungcuso3luongyen

Power of hydrogen ion pH-

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 ab = 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

a)

pH = – log Ka + 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 Ka + 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 ab = 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

b)

pOH = – log Kb + log [salt ] /[base]

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

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

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

[log ab = 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

c)

pH = – log Ka + 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 Ka + 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 ab = 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