Hi again!

Question

I have a new question, Can you help me?
Consider 50.0 mL of a solution of a weak acid HA (Ka = 1.00.E-6), which has a pH of 4.000. What volume of water must be added to make the pH = 5.000?
My Calculations:
To calculate the concentration of x, I take the pH value ->
pH = 4,000=> x=[H+]= 1.E-4=0.0001M=
x=1.E-4M

Ka=(1,0.E-4 )^2/(x-1*E-4)=(1,00.E-6); solving x = 0,0101M

To calculate the quantity of moles in the solution, I do the following:
(50,0*10-3L)* (0,0101M) = 5,05*10-4 mol HA.
I don´t know what to do next:-(

Answer 1

Go over the approximate solution solution I gave you in the previous question. You can improve the values you get by substituting and solving the quadratic equation,
Ka = x^2 / c - x)
The approximate solution is probably adequate. Check the method here:
http://Jiskha.com/display.cgi?id=1218967283.1219017604