SIGNME Question 1 HA is a weak acid. Its ionization constant, K,is 33 x 10-13. C
ID: 1045313 • Letter: S
Question
SIGNME Question 1 HA is a weak acid. Its ionization constant, K,is 33 x 10-13. Calculate the pH of an aqueous solution with an initial NaA concentration of 0.051 M Question 2 We place 0.134 mol of a weak acid, HA, in enough water to produce 1.00 L of solution. The final pH of the solution is 1.13. Calculate the ionization contant, Ka of HA. Question 3: We place 0.542 mol of a weak acid, HA, and 12.3 g of NaOH in enough water to produce 1.00 L of solution. The final pH of this solution is 4.49. Calculate the ionization constant, Kg, of HAExplanation / Answer
NaA (aq) ----------------> Na^+ (aq) + A^- (aq)
0.051M 0.051M
A^- (aq) + H2O ----------------> HA + OH^-
I 0.051 0 0
C -x +x +x
E 0.051-x +x +x
Kb = Kw/Ka
= 1*10^-14/3.3*10^-13 = 0.03
Kb = [HA][OH^-]/[A^-]
0.03 = x*x/0.051-x
0.03*(0.051-x) = x^2
x = 0.0269
[OH^-] = x = 0.0269M
POH = -log[OH^-]
= -log0.0269
= 1.57
PH = 14-POH
= 14-1.57 = 12.43
2 PH = 1.13
-log[H^+] = 1.13
[H^+] = 10^-1.13 = 0.074M
HA(aq) ------------------> H^+ (aq) + A^- (aq)
I 0. 134 0 0
C -0.074 0.074 0.074
E 0.06 0.074 0.074
Ka = [H^+][A^-]/[HA]
= 0.074*0.074/0.06 = 0.0912 = 9.12*10^-2
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