Question

Calculate [H+] and [CHCl2COO-] in a solution of 0.01 M HCl and 0.01 M CHCl2COOH. (Take Ka = 2.55*10^-2)

Answer 1

Given,

Concentration of strong acid, HCl = 0.01 M

Concentration of weak acid, CHCl₂COOH = 0.01 M

Acid dissociation constant, Kₐ = 2.55 × 10⁻²

Both  the strong and weak acids dissociate as follows.

                      HCl      →      H⁺      +    Cl⁻

                    0.01 M         0.01 M     0.01 M

                 

                      CHCl₂COOH ⇔ CHCl₂COO⁻ + H⁺

At time t=0        0.01 M                    -                 -

At time t=t         (0.01-x)                   x               (x+0.01) , {here [H⁺] includes the value from both dissociation of strong and weak acids}

∴ Kₐ = $\frac{x(x+0.01)}{(0.01-x)}$

⇒ 0.0255(0.01-x) = x² + 0.01x

⇒ x² + 0.0355x - 0.000255 = 0

Using the binomial root formula,

x =  $\frac{-0.0355±[tex]\sqrt{0.04775}$}{2}[/tex] = 1.1 × 10⁻² = 0.011

∴ [CHCl₂COO⁻] = 0.011

∴ Net [H⁺] = x + 0.01 =  0.011 + 0.01 =0.021

Answer 2

Answer:

Refer the image attached

Explanation: