# The equilibrium constant (K_(c)) for the reaction of a weak acid HA with strong base NaOHis 10 at 25^(@)C. Which of the following are correc

506 views

closed
The equilibrium constant (K_(c)) for the reaction of a weak acid HA with strong base NaOHis 10 at 25^(@)C. Which of the following are correct deduction
A. The ionization constant K_(a) at 25^(@)C is 10^(-5)
B. pH of a 0.01 M aqueous solution of HA at 25^(@)C will be 3.5
C. pH of a 0.01 M aqueous solution of NaA at 25^(@)C will be 9.
D. If K_(b) of weak base BOH is 10^(-4) at 25^(@)C, equilibrium constant for neutralization of HA with BOH at 25^(@)C will be 10^(5)

by (65.9k points)
selected

underset(("Weak"))(HA) + NaOH hArr NaA + H_(2)O , K = 10^(9)
K = 1/(K_(h)) hArr = 10^(-9)
A) K_(h) = (K_(w))/(K_(a)) or K_(a) = (K_(w))/(K_(h)) = 10^(-5)
B) pH of 0.01 M HA = 1/2 pK_(a) - 1/2 "log" C = 3.5
C) pH of 0.1 M aq.NaA = 7 + 1/2 pK_(a) + 1/2 "log" C = 9
K = (1)/(K_(h)) = (K_(a) xx K_(b))/(K_(w)) = (10^(-5) xx 10^(-4))/(10^(-14)) = 10^(5)