Question

what the ph of 0.109 M hydrogen chromate ion (HCrO4^-)

Answer 1

Answer - We are given, [HCrO₄^-] = 0.109 M ,

We know Ka2 for the H₂CrO₄ = 3.2*10^-7

we need to put ICE table for calculating the [H₃O⁺]

    HCrO₄^- + H₂O ------> H₃O⁺ + CrO₄^2-

I 0.109                          0           0

C   -x                            +x          +x

E 0.109-x                       +x          +x

Ka = [H₃O⁺] [CrO₄^2-] / [HCrO₄^-]

3.2*10^-7 = x*x /(0.005-x)

3.2*10^-7 *(0.109-x) = x²

The Ka value is too small, so we can neglect the x in the 0.109-x.

3.2*10^-7*0.109 =x²

x = 0.000187 M

so, x = [H₃O⁺] = 0.000187 M

so, pH = -log [H₃O⁺]

           = -log 0.000187 M

           = 3.73

So answer for this question is 3.73