Calculate the pH of 0.65 M H2SO4 (sulfuric acid) solution? (Ka2: 1.1x10-2)

Explanation:

.65 M H₂SO₄

     H₂SO₄ ⇄ 2H⁺ + SO₄⁻²

  C(1-x)             2xC         xC

(2xC)². xC /C( 1- x )  = 1.1 x 10⁻²

4 x³C² =  1.1 x 10⁻²

4 . x³ .65² = 1.1 x 10⁻²

x³ =   .0065 = 6.5 x 10⁻³

x =      1.867 x 10⁻¹

H⁺ ion concentration =

2xC = 2 x 1.867 x 10⁻¹ x .65

=  .2427

pH = - log ( .2427 )

= 0 .615 .    

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