Consider 54.0 mL of a solution of weak acid HA (Ka = 1.00 10-6), which has a pH of 4.200. What volume of water must be added to make the pH = 5.500?
I found HA for both and then I used C1V1 = C2V2 plugging in the concentrations of HA in C1 and C2 and 0.054 for V1, but I'm still not getting the right answer. Please help!
To solve this problem, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, the initial pH is 4.200, and we want to adjust it to 5.500 by adding water. Let's assume the initial volume of the solution is V0 mL, and the final volume after adding water is V1 mL.
We need to find the concentration of the weak acid and its conjugate base in terms of [HA] and [A-].
Given:
Initial pH = 4.200
pKa = -log(Ka) = -log(1.00 x 10^(-6)) = 6
Using the Henderson-Hasselbalch equation, we can write:
4.200 = 6 + log([A-]/[HA])
Now, let's solve for [A-]/[HA]:
log([A-]/[HA]) = 4.200 - 6
[A-]/[HA] = 10^(4.200 - 6)
[A-]/[HA] = 10^(-1.800)
Since the ratio of [A-]/[HA] is equal to the ratio of the volumes in which they exist, we can say:
[V(A-)]/[V(HA)] = 10^(-1.800)
The initial volume of the solution is 54.0 mL. At equilibrium, the final volume is the sum of this initial volume and the volume of water added, V1.
So, we can write:
[V(A-)]/[V(HA) + V(water)] = 10^(-1.800)
Substituting [V(A-)] = V1 mL and [V(HA)] = 54.0 mL in the equation, we get:
V1 / (54.0 + V1) = 10^(-1.800)
Now, we can solve this equation for V1 to find the volume of water needed to achieve the desired pH.
Unfortunately, without the specific values for [HA] and [A-], we cannot solve for V1 accurately. Please check if you have provided all the necessary information or recheck your calculations.