Calculate the pH of a 9.06×10-3 M solution of H2SO4. (Ka = 0.0120 for HSO4-)

Calculate the pH of a 9.06×10^-3 M solution of H₂SO₄.
(K_a = 0.0120 for HSO₄^-)

Expert Solution

H2SO4(aq) --------------> H^+ (aq) + HSO4^- (aq)

9.06*10^-3M 9.06*10^-3M

HSO4^- (aq) -----------------> H^+ (aq) + SO4^2- (aq)

I 9.06*10^-3 0 0

C -x +x +x

E 9.06*10^-3 -x +x +x

Ka = [H^+][SO4^2-]/[HSO4^-]

0.012 = x*x/9.06*10^-3 -x

0.012*(0.00906-x) = x^2

x = 0.006

[H^+] = x = 0.006M

PH = -log[H^+]

= -log0.06

= 1.2218

queen_honey_blossom answered 2 years ago

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