Question

Find the pH and [SO42-] of a 0.0050 M sulfuric acid solution

Answer 1

Answer – We are given, weak base concentration [H₂SO₄]= 0.0050 M

The first dissociation of the H₂SO₄ is complete , so it is

[H₂SO₄]= [HSO₄^-] = [H₃O⁺] = 0.0050

Ka for HSO₄^- = 1.2*10^-2

we know sulfuric acid second dissociate is not completely, so we need to put ICE table for calculating the [H₃O⁺]

    HSO₄^- + H₂O ------> H₃O⁺ + SO₄^2-

I 0.005                          0           0

C -x                              +x          +x

E 0.005-x                      +x          +x

Ka = [H₃O⁺] [SO₄^2-] / [HSO₄^-]

1.2*10^-2 = x*x /(0.005-x)

1.2*10^-2 *(0.005-x) = x²

Now we need to set up quadratic equation

6*10^-5 - 0.012x = x²

x^{2 +} 0.012x – 6.0*10^-5 = 0

a =1 , b = 0.012 , c = 6.0*10^-5

Using the quadratic equation

x = -b +/- √b²-4a*c / 2a

Plugging the value in this formula

x = 0.00380 M

so, x = [H₃O⁺] = 0.00380 M

[SO₄^2-] = 0.00380 M

Total [H₃O⁺] = 0.0050+0.00380 = 0.0088 M

so, pH = -log [H₃O⁺]

           = -log 0.0088 M

           = 2.05