Chemistry/pH Weak Acid
posted by Mary .
Hi again!
I have a new question, Can you help me?
Consider 50.0 mL of a solution of a weak acid HA (Ka = 1.00.E6), 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.E4=0.0001M=
x=1.E4M
The equilibrium reaction is.
HA H+ + A
Initial x ~0 0
Final x1.E4 x1.E4 x1.E4
Ka=(1,00.E4)^2/(x1*E4)=1,00.E6;
solving x = 0,0101M
To calculate the quantity of moles in the solution, I do the following:
(50,0*103L)* (0,0101M) = 5,05*104 mol HA.
I donĀ“t know what to do now!
Thanks!

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 = ______? 
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/(x1*10^(4) )=1,00*10^(6)....I got c=0,0101M
:( 
where did you get c = 1.00x10^2 = 0.0100 M from ?