14. Consider a solution of formic acid (HCHO2) and sodium formate (NaCHO2), at pH=3.70. After the addition of 0.015 moles of HNO3, the pH decreases by 0.12. What are the initial molarities of HCHO2 and CHO2-, if the Ka of HCHO2 is 1.8x10-4?

Equilibrium system:

HCHO2 (aq) H+ (aq) + CHO2- (aq)

x = [HCHO2], y = [CHO2-], 0.912x = y

Knowing the reaction that occurs, the stoichiometric relationship must be determined:

CHO2- (aq) + H+ (aq) HCHO2 (aq)

CHO2- H+ HCHO2 (aq)
INITIAL
0.912x mol
0.015 mol
x mol
CHANGE
-0.015 mol
-0.015 mol
+0.015 mol
FINAL
(0.912x - 0.015) mol
0 mol
(x + 0.015) mol

What happens next?

a) By Le Chatelier's principle, the system favours the acid
b) Because there is no H+, the pH is also 0
c) A new equilibrium forms to satisfy the Ka
d) The buffer capacity has been exceeded and fails