In: Chemistry
Determine the pH of a 0.115M sodium citrate (NAC6H7O7) solution. The Ka for citric acid (HC6H7O7) is 7.4X10^-4
NaC6H7O7(aq) ---------------> Na^+ (aq) + C6H7O7^- (aq)
0.115M 0.115M
C6H7O7^- (aq) + H2O(l) -------------------> HC6H7O7(aq) + OH^- (aq)
I 0.115 0 0
C -x +x +x
E 0.115-x +x +x
Kb = Kw/Ka
= 1*10^-14/7.4*10^-4 = 1.35*10^-11
Kb = [HC6H7O7][OH^-]/[C6H7O7^-]
1.35*10^-11 = x*x/0.115-x
1.35*10^-11*(0.115-x) = x^2
x = 1.25*10^-6
[OH^-] = x = 1.25*10^-6M
POH = -log[OH^-]
= -log1.25*10^-6
= 5.9030
PH = 14-POH
= 14-5.9030 = 8.097 >>>>answer