Ionic EquilibriumMedium
Question
The pH of 10-8 M solution of HCl in water is :
Options
A.8
B.- 8
C.between 7 and 8
D.between 6 and 7
Solution
H2O ⇋ H+ + OH-
(10-8 + a) a
a is the concentration of [H+] furnished by the dissociation of H2O.
Kw = [H+][OH-]
10-14 = (10-8 + a)(a)
a = 0.95 × 10-7
[H+] = 10-8 + 0.95 × 10-7
= 1.05 × 10-7
pH = - log10 1.05 × 10-7
= 6.9788
(10-8 + a) a
a is the concentration of [H+] furnished by the dissociation of H2O.
Kw = [H+][OH-]
10-14 = (10-8 + a)(a)
a = 0.95 × 10-7
[H+] = 10-8 + 0.95 × 10-7
= 1.05 × 10-7
pH = - log10 1.05 × 10-7
= 6.9788
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
For the reaction H2(g) + I2 (g) ⇋ 2HI(g) The equilibrium constant Kp changes with :...Which of the following the weakest base is :...In the following reaction H2CO3 and C6H5NH2 respectively act as :-H2CO3 + C6H5NH2 ⇋ HCO3- + C6H5NH3+...Ka for HCN is 5 × 10-10 at 25oC. For maintaining a constant pH of 9, the volume of 5 M KCN solution required to be ...What is the pH of solution made by mixing equal volumes of 0.1 N-H2SO4, 0.1 N-HNO3, 0.1 N-HCl?...