Question

Percentage ionization of water at a certain temperature is 3.6×10^-7 . Calculate Kw and ph

Answer 1

Hii dear,

Given -

Actual ionization of water at certain temperature = Percentage ionisation/100 = 3.6×10^-7 /100 = 3.6×10^-9

Solution=

Ionization of water occurs as follows

                2H20 ----> OH- + H3O+

Initial moles     1          0     0

At Equilibrium 1-3.6×10^-9  3.6×10^-9 (for both)

Now

Kw = [OH-][H30+]

Kw = 3.6×10^-9 × 3.6×10^-9

Kw = 12.96 × 10^-18

And

pH = log[H30+]

pH = log 3.6×10^-9

pH = 8.4

Hope that solved your doubt.

Answer 2

Explanation:

Dear Student,

Suppose volume of water = 1 L

Density of water = 1 g/ml

So, mass of 1 L = 1000 g

Number of moles of water = 1000/18 = 55.5 moles

So, number of moles present = 55.5 mol/L

The ionization of water is:

H2O ---> H+ + OH-

So, [H+] = [OH-]

Percent ionization = ([H+]/[H2O])×100

3.6×10^−7=([H+]/55.5)×100

[H+] = 199.8×10^−9 = 2×10^−7 M

[OH-] = [H+] = 2*10^-7 M

Kw = [H+][OH-] = (2*10^-7)^2 = 4*10^-14

pH = -log[H+] = -log(2*10^-7) =7– log 2

= 7– 0.3010 = 6.6990

pH = 6.6990