Calculate the volume of .300 M NaOH needed to prepare a pH=3.50 buffer with 250ml of .075M HF solution. pKa (HF) = 3.15
You have 250 mL x 0.075M = 18.75 mmoles HF.
........HF + NaOH ==> NaF + H2O
initial.18.75..0.......0......0
add.............x................
change..-x.....-x......+x.....+x
equil..18.75....0........x.....x
pH = pKa + log [(base)/(acid)]
3.50 = 3.15 + log(x/18.75-x)
Solve for x which gives millimoles, then
mmoles = mL x M
You know mmoles and M, solve for mL of the 0.3M NaOH to add.
To calculate the volume of 0.300 M NaOH needed to prepare a pH=3.50 buffer with 250 ml of 0.075 M HF solution, we need to use the Henderson-Hasselbalch equation for a buffer system:
pH = pKa + log ([A-]/[HA])
Where pH is the desired pH of the buffer, pKa is the dissociation constant of the acid (in this case, HF), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
First, let's determine the concentration of the acid and the conjugate base in the buffer solution.
Given:
pKa (HF) = 3.15
pH = 3.50
[HF] = 0.075 M
To find [A-], we can use the relationship:
[A-] = [HF] * (10^(pH - pKa))
[A-] = 0.075 M * (10^(3.50 - 3.15))
[A-] = 0.075 M * (10^0.35)
[A-] = 0.075 M * 2.2387211
[A-] = 0.1679048 M
Now, since a buffer solution requires an equal concentration of the conjugate base and the acid, the concentration of [HA] is also 0.1679048 M.
To calculate the volume of 0.300 M NaOH needed, we can use the equation:
(0.1679048 M)(V) = (0.300 M)(V + 250 mL)
Simplifying:
0.1679048V = 0.3(V + 0.25)
0.1679048V = 0.3V + 0.075
0.3V - 0.1679048V = 0.075
0.1320952V ≈ 0.075
V ≈ 0.075 / 0.1320952
V ≈ 0.568 mL
Therefore, approximately 0.568 mL of 0.300 M NaOH is needed to prepare a pH=3.50 buffer with 250 mL of 0.075 M HF solution.