Posted by Mary on Sunday, August 17, 2008 at 6:07am.
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
Ka=(1,0.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 next:(

Chemistry/pH Weak Acid  GK, Sunday, August 17, 2008 at 8:08pm
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