Consider 67.0 mL of a solution of weak acid HA (Ka = 1.00 10-6), which has a pH of 4.500. What volume of water must be added to make the pH = 5.000?
HA ==> H^+ + A^-
Solve this equation to determine the (HA) at pH 4.50.
As a separate problem, plug in pH 5.000 and determine the (HA) for that value.
Then use the dilution formula using 67.0 mL of the first to determine how much to dilute it.
c1v1 = c2v2
Post your work if you get stuck.
Thank you very much.
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the concentration ratio of its conjugate acid-base pair. The equation is given by:
pH = pKa + log([A-]/[HA])
Where pH is the given pH value, 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.
First, let's find the initial concentration of the weak acid HA. Since we are given the pH, we can use the equation pH = -log[H+], where [H+] is the concentration of the hydrogen ions.
Given pH = 4.500, we can calculate [H+] as follows:
[H+] = 10^(-pH) = 10^(-4.500)
Next, we need to convert the concentration of hydrogen ions to the concentration of the acid HA. Since the acid is weak, we can assume that it dissociates very little, so we can approximate the concentration of the acid as being equal to the concentration of hydrogen ions:
[HA] = [H+]
Now, let's substitute the known values into the Henderson-Hasselbalch equation:
4.500 = -log(1.00x10^(-6)) + log([A-]/[HA])
Simplifying the equation:
4.500 = 6 - log[HA] + log[A-]
Rearranging the equation:
log[HA] - log[A-] = 1.5
Since the logarithms have the same base, we can combine them into a single logarithm:
log([HA]/[A-]) = 1.5
Now, let's solve for the concentration ratio [HA]/[A-]:
[HA]/[A-] = 10^(1.5)
Next, we need to find the initial molarities of [HA] and [A-]. We know that [HA] = [H+] = 10^(-4.500). Since the concentration of [A-] is equal to the initial concentration of the added water since it does not react with the weak acid, we will denote it as [A-]initial.
[A-]initial = 0 M (since it is pure water)
Now, let's solve for [HA]initial:
[HA]initial / [A-]initial = 10^(1.5)
[HA]initial / 0 = 10^(1.5)
[HA]initial = infinity
Since the initial concentration of [HA] is infinite, it means that the acid is completely undissociated, and in such a case, we cannot use the Henderson-Hasselbalch equation.
Therefore, no matter how much water we add, the pH will remain at 4.500.