A buffer solution is made by dissolving 0.45
moles of a weak acid (HA) and 0.23 moles of
KOH into 720 mL of solution. What is the
pH of this buffer? Ka = 6.2 × 10−6
for HA.
Answer in units of pH.
please i need the answerf
To determine the pH of the buffer solution, we can use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log([A-]/[HA])
Where:
- pH is the measure of the acidity or basicity of a solution.
- pKa is the negative logarithm of the acid dissociation constant (Ka) for the weak acid (HA).
- [A-] is the concentration of the conjugate base (A-) of the weak acid.
- [HA] is the concentration of the weak acid (HA).
In this case, the weak acid being used is HA, and its concentration is given as 0.45 moles (since we're given the number of moles). The volume of the solution is given as 720 mL, but we need to convert it to liters by dividing it by 1000:
Volume = 720 mL = 720/1000 = 0.72 L
To find the concentration of [A-], we need to consider that KOH reacts with the weak acid HA to form its conjugate base A-. The mole ratio between HA and KOH is 1:1, which means that the concentration of [A-] is the same as the concentration of KOH, given as 0.23 moles.
Now we can calculate the pH using the Henderson-Hasselbalch equation. We need to start by finding the pKa value. The pKa can be obtained by taking the negative logarithm of the Ka value given for HA:
pKa = -log10(Ka) = -log10(6.2 × 10^-6) ≈ 5.21
Next, substitute the known values into the equation:
pH = 5.21 + log10(0.23/0.45)
Now, calculate the ratio of [A-]/[HA]:
[A-]/[HA] = 0.23/0.45 ≈ 0.511
Finally, calculate the pH:
pH ≈ 5.21 + log10(0.511)
Using a calculator, find the logarithm of 0.511:
log10(0.511) ≈ -0.290
Now, substitute this value back into the equation:
pH ≈ 5.21 - 0.290
Calculating this expression, we get:
pH ≈ 4.92
Therefore, the pH of the buffer solution is approximately 4.92.