Question

How do you calculate the pH of 1.0 Moles of HClO, HCO2H, and KHCO2?

Answer 1

1. HClO + H2O ⇌ H3O^+ + ClO^- Ka= 3.5 * 10^ -8

Initial moles: 1.0 mole - -

At equilibrium:   1.0-x mole x mole x mole

Assume Volume to be 1 litre, then

Concentration: (1.0-x) M x M x M

Now, Ka= [H3O^+] [ClO^-] / [HClO]

=> 3.5*10^-8 = x*x/ (1.0-x)

As Ka value is very small, we can neglect x in the denominator,

=> 3.5* 10^-8 = x^2

=> x = 1.87 * 10^-4

So, [H3O^+] = 1.87 * 10^-4 M

pH= -log(1.87 * 10^-4)

=> pH= 3.73

2. HCO2H + H2O ⇌ H3O^+ + HCOO^- Ka= 1.8 ×10-4

Initial moles: 1.0 mole - -

At equilibrium:   1.0-x mole x mole x mole

Assume Volume to be 1 litre, then

Concentration: (1.0-x) M x M x M

Now, Ka= [H3O^+] [HCOO^-] / [HCO2H]

=> 1.8*10^-4 = x*x/ (1.0-x)

As Ka value is very small, we can neglect x in the denominator,

=> x^2 = 1.8*10^-4

=> x= 1.34*10^-2

So, [H3O^+] = 1.34*10^-2 M

pH= -log(1.34*10^-2)

=> pH = 1.87

3.   KHCO2 + H2O ⇌ H3O^+ + HCO2^- Ka= 1.78x10-4

Initial moles: 1.0 mole - -

At equilibrium:   1.0-x mole x mole x mole

Assume Volume to be 1 litre, then

Concentration: (1.0-x) M x M x M

Now, Ka= [H3O^+] [HCO2^-] / [KHCO2]

=>1.78*10^-4 = x*x/ (1.0-x)

As Ka value is very small, we can neglect x in the denominator,

=> 1.78* 10^-4 = x^2

=> x = 1.33 * 10^-2

So, [H3O^+] = 1.33 * 10^-2 M

=> pH= -log(1.33 * 10^-2)

=> pH= 1.88