In: Chemistry
HNO2(aq) -------------> H^+ (aq) + NO2^- (aq)
I 0.036 0 0
C -x +x +x
E 0.036-x +x +x
Ka = [H^+][NO2^-]/[HNO2]
4.5*10^-4 = x*x/0.036-x
4.5*10^-4 *(0.036-x) = x^2
x = 0.0038
[H^+] = x = 0.0038M
PH = -log[H^+]
= -log0.0038
= 2.42 >>>>answer