In: Chemistry
Calculate the pH and the equilibrium
concentrations of
H2C6H5O7-,
HC6H5O72-
and
C6H5O73-
in a 0.0858 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 0.0858 0 0
C -x +x +x
E 0.0858-x +x +x
Ka1 = [H^+][H2C6H5O7^-]/[H3C6H5O7]
7.4*10^-3 = x*x/0.0858-x
7.4*10^-3 *(0.0858-x) = x^2
x = 0.0217
[H^+] = x = 0.0217M
[H2C6H5O7^-] = x = 0.0217M
[H3C6H5O7] = 0.0858-x = 0.0858-0.0217 = 0.0641M
PH = -log[H^+]
= -log0.0217 = 1.6635 >>>answer
H2C6H5O7^- (aq) ---------------> H^+ (aq) + HC6H5O7^2- (aq)
ka2 = [H^+][HC6H5O7^2-]/[H2C6H5O7^-]
1.7*10^-5 = 0.0217*[HC6H5O7^2-]/0.0217
[HC6H5O7^2-] = 1.7*10^-5 M >>>answer
HC6H5O7^2- (aq) -------------> C6H5O7^3-(aq) + H^+ (aq)
ka3 = [C6H5O7^3-][H^+]/[[HC6H5O7^2-]
4*10^-7 = [C6H5O7^3-]*0.0217/1.7*10^-5
[C6H5O7^3-] = 4*10^-7 *1.7*10^-5/0.0217 = 3.13*10^-10 M >>>answer