Question
The pH of a 0.175 M aqueous solution of a weak acid is 3.52. What is Ka for this acid ?
Answers
bobpursley
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.
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.
Sammie
thank you so much!