Question
If a 0.200M solution of an acid, HX, has a pH=1.00, The Ka=
a. 2 x 10^-1
b. 2 x 10^-2
c. 1
d. 1 x 10^-1
e. 1 x 10^-3
This is what I did:
HX <===> H + X
10^-1=[H]
0.1=[H]
[H]=[X]
Ka = (H)(X)/HX
Ka = (0.1)(0.1)/0.200
Ka = 0.05
a. 2 x 10^-1
b. 2 x 10^-2
c. 1
d. 1 x 10^-1
e. 1 x 10^-3
This is what I did:
HX <===> H + X
10^-1=[H]
0.1=[H]
[H]=[X]
Ka = (H)(X)/HX
Ka = (0.1)(0.1)/0.200
Ka = 0.05
Answers
bobpursley
HX should = .2-.1, right?
so Ka=.01/.1=.1
check my thinking
so Ka=.01/.1=.1
check my thinking
Anonymous
Seems correct to me