The pH of a 0.175 M aqueous solution of a weak acid is 3.52. What is Ka for this acid ?

Answer 1

Ka=[H+][attached-]/(.175-x)
or

ka= x^2/(.175-x)

but pH=-logx, or x=10^-3.52

so x^2=(3.04E-4)^2=about 9E-8 check that

Ka= (9E-8)/(.175-.000304)

ka=9E-8/.175 =1.73E-3

check all this, I did most of it in my head.

Answer 2

thank you so much!