Question

What are the concentrations of HSO₄⁻, SO₄²⁻, and H⁺ in a 0.37 M KHSO₄ solution?
(Hint: H₂SO₄ is a strong acid; K_a for HSO₄⁻ = 1.3

×

10⁻².)

Answer 1

We Have

Ka for HSO₄^- = 1.3 $\times$ 10^-2 = 0.013

[HSO₄^-] = 0.37 M

Now lets see the ICE table

HSO₄^-   $\rightleftharpoons$ H⁺ + SO₄^2-  

Initial 0.37    0 0

Change -x +x +x

Equilibrium 0.37-x x x

Equilibrium constant for above reaction is

Ka = [H⁺][SO₄^2-]/[HSO₄^-] = (x)(x) /(0.37-x) = x² / (0.37-x)

x² / (0.37-x) = Ka = 1.3 $\times$ 10^-2 = 0.013

as x is too small than 0.37 , 0.37-x= 0.37

then

x² /0.37 = 0.013

x² = 0.00481

taking square roots on both the sides we get

x = 0.069 (or approx 0.07 M)

So

[H⁺] = x = 0.07 M

[SO₄^2-] = x= 0.07 M

[HSO₄^-] = 0.37 -x = 0.37 - 0.07 = 0.30 M