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
The equilibrium reaction is.

HA H+ + A-
Initial x ~0 0
Final x-1.E-4 x-1.E-4 x-1.E-4

Ka=(1,00.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 now!

Thanks!

Answer 1

For a pH = 4, [H+] = antilog(-4.000) = 1.00x10^-4
Ka = [H+][A-] / [HA]
Let [H+] = [A-] = x
Ka = x^2 / (c - x) , where c = molar concentration of HA before dissociation. if c is much larger than x, we can simplify the expression to:
Ka = x^2 / c
(1.00x10^-6)c = 1.00x10^-8
c = 1.00x10^-2 = 0.0100 M
If we want a pH of 5, we can calculate a new value for c (as we did above) which is lower than the previous value of 0.0100 M.
Calculate c for pH = 5, then use:
(0.0100M)(50.0mL) = (c)(V)
substitute the 2nd value of c and solve for V. Then volume added =
V - 50.0mL = ______?

Answer 2

but the rule of 5%?
0,0100/(1.00x10^-4) =100%
The aproximation you do is not valid...or?
Ka= (1,0*10^(-4))^2/(x-1*10^(-4) )=1,00*10^(-6)....I got c=0,0101M
:-(

Answer 3

where did you get c = 1.00x10^-2 = 0.0100 M from ?