A buffer solution is made by dissolving 0.45 moles of a weak acid (HA) and 0.23 moles of KOH into 680 mL of solution. What is the pH of this buffer? Ka = 2.4×10−6 for HA.
.......KOH + HA ==> KA + H2O
I.....0.23..0.45....0....0
C....-0.23.-0.23...0.23..0.23
E......0....0.22...0.23
(KA) = 0.23 mols/0.680 L = ?M
(HA) = 0.22 mols/0.680 L = ?M
pH = pKa + log (base)/(acid)
pH = ?
5.63
To find the pH of the buffer solution, we need to determine the concentration of HA and its conjugate base A- in the solution. Then, we can use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log([A-]/[HA])
Given the number of moles of HA and KOH, we can calculate the concentration of HA and A- in the buffer solution.
Step 1: Calculate the volume of the solution in liters.
The volume given is 680 mL, which is equivalent to 0.680 L.
Step 2: Calculate the concentration of HA.
The concentration of HA is given by dividing the number of moles of HA by the volume of the solution:
[HA] = moles of HA / volume (L)
[HA] = 0.45 moles / 0.680 L = 0.662 M (rounded to three decimal places)
Step 3: Calculate the concentration of A- (conjugate base).
Since KOH is a strong base, it reacts completely with the weak acid HA to form the conjugate base A-. Therefore, the concentration of A- is equal to the moles of KOH divided by the volume of the solution:
[A-] = moles of KOH / volume (L)
[A-] = 0.23 moles / 0.680 L = 0.338 M (rounded to three decimal places)
Step 4: Calculate the pH using the Henderson-Hasselbalch equation.
We are given the Ka value for HA, which is 2.4 × 10^(-6). Since we know the value of pKa is equal to the negative logarithm of Ka (pKa = -log(Ka)), we can calculate pKa as follows:
pKa = -log(2.4 × 10^(-6)) ≈ 5.62 (rounded to two decimal places)
Now, plug the values into the Henderson-Hasselbalch equation:
pH = 5.62 + log(0.338/0.662) ≈ 5.28 (rounded to two decimal places)
Therefore, the pH of the buffer solution is approximately 5.28.