Question

Consider 54.0 mL of a solution of weak acid HA (Ka = 10^-6), which has a pH of 4.140. What volume of water must be added to make the pH = 5.000?

[HA]=((10^-4.14)^2)/(10^-6)
= 0.0052480746

0.0052480746 x 54 = c x v

c= ((10^-5)^2)/(10^-6)
= 10^-4

v= (0.0052480746 x 54)/(10^-4)

But my web assign says the answer is wrong, how?

Answer 1

let the final total vol = X

.054L * [H]init / X liters = [H]final

[H]init = 10^-4.14 mol/L
[H]fin = 10^-5 mol/L

by algebra, X=.3912 Liters final vol.

391.2 mL - 54 mL = 337.2 mL to add.

This may be what you are looking for, but it glosses over a lot of detail.