In: Chemistry
What are [HCOOH], [H3O+], [HCOO–], [OH–] and pH of
a) 0.300 M HCOOH,
b) 0.005 M HCOOH solutions? (Ka = 1.8·10–4)
HCOOH(aq) + H2O(l) H3O+(aq) + HCOO–(aq)
HCOOH(aq) + H2O(l)--------------> H3O+(aq) + HCOO–(aq)
I 0.3 0 0
C -x +x +x
E 0.3-x +x +x
Ka = [H3O^+][HCOO^-]/[HCOOH]
1.8*10^-4 = x*x/0.3-x
1.8*10^-4 *(0.3-x) = x^2
x= 0.00726
[HCOOH] = 0.3-x = 0.3-0.00726 = 0.29274M
[H3O^+] = x = 0.00726M
[HCOO^-] = x = 0.00726M
[OH-] = Kw/[H3O^+]
= 1*10^-14/0.00726 = 1.37*10^-12M
PH = -log[H3O+]
= -log1.37*10^-12 = 11.86 >>>>answer
b.
HCOOH(aq) + H2O(l)--------------> H3O+(aq) + HCOO–(aq)
I 0.005 0 0
C -x +x +x
E 0.005-x +x +x
Ka = [H3O^+][HCOO^-]/[HCOOH]
1.8*10^-4 = x*x/0.005-x
1.8*10^-4 *(0.005-x) = x^2
x= 0.000863
[HCOOH] = 0.005-x = 0.005-0.000863 = 0.004137M
[H3O^+] = x = 0.000863M
[HCOO^-] = x = 0.000863M
[OH-] = Kw/[H3O^+]
= 1*10^-14/0.000863 = 1.158*10^-11M
PH = -log[H3O+]
= -log0.000863 = 3.0639 >>>>answer