A .050M solution of a weak acid HA has [H3O+] =3.77 x 10^ -4. What is the Ka for this acid?

Question

A. 7.5 x 10^ -3

B. 7.0 x 10^ -8

C. 2.8 x 10^-6

D. 2.6 x 10^-11

Answer 1

AN EDIT:  the other answerer copied & pasted my answer, 3 days after I posted mine. Presenting it as his is plagiarism. Others have registered reports of his abuse to "Yahoo Answers". would you please report him as well, by going to the bottom of HIS answer & clicking on "report abuse".  this is the 10th time he has presented one of  my answers as his.

======================================...

HA -->  H+  &  A-  

0.050 -->x   &  x  .... 1:1 ratio

Ka = [ H+]  [A- ] / [HA]

Ka = [ 3.77 x 10^ -4]  [3.77 x 10^ -4 ] / [0.050]

your answer is: Ka = 2.8e-6  ... "C"