Calculate the pH and the equilibrium concentrations of H2C6H5O7-, HC6H5O72- and C6H5O73- in a 0.1840 M aqueous citric acid solution. For H3C6H5O7, Ka1 = 7.4×10-3, Ka2 = 1.7×10-5, and Ka3 = 4.0×10-7
pH =
[H2C6H5O7-] = __M
[HC6H5O72-] =__ M
[C6H5O73-] = __M
H3C6H5O7 (aq) ----------------> H^+ (aq) + H2C6H5O7^- (aq)
I 0.184 0 0
C -x +x +x
E 0.184-x +x +x
ka1 = [H^+][H2C6H5O7^-]/[H3C6H5O7]
7.4*10^-3 = x*x/(0.184-x)
7.4*10^-3*(0.184-x) = x^2
x = 0.0334
[H^+] = x = 0.0334M
[H2C6H5O7^-] = x = 0.0334M
PH = -log[H^+]
= -log0.0334
= 1.4762
H2C6H5O7^-(aq) ---------------> H^+ (aq) + HC6H5O7^2-(aq)
Ka2 = [H^+][HC6H5O7^2-]/[H2C6H5O7^-]
1.7*10^-5 = 0.0334[HC6H5O7^2-]/0.0334
[HC6H5O7^2-] = 1.7*10^-5 M
HC6H5O7^2-(aq) ---------------> H^+ (aq) + C6H5O7^3-(aq)
Ka3 = [H^+][C6H5O7^3-]/[HC6H5O7^2-]
4*10^-7 = 0.0334[C6H5O7^3-]/1.7*10^-5
[C6H5O7^3-] = 4*10^-7*1.7*10^-5/0.0334 = 2*10^-10M
pH =1.4762
[H2C6H5O7-] = 0.0334M
[HC6H5O72-] =1.7*10^-5M
[C6H5O73-] = 2*10^-10 M
Calculate the pH and the equilibrium concentrations of H2C6H5O7-, HC6H5O72- and C6H5O73- in a 0.1840 M...
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