Question

4. The stockroom contains 1.0 M NaAc (Sodium Acetate), 1.0 M HAc (acetic Acid), distilled water and strong acids and bases. You need to create a buffer solution that has a pH of 4.75 such that when 1.00 mL of 10.0 M HCl is added to 100.00 mL of your buffer, the resulting pH is 3.75 (+/- 0.1). What concentrations of HAc and NaAc do you need to create the buffer solution? Show your calculations and also write out steps on how you "virtually" performed the titration.

Answer 1

Answer:

Concentrations of HAc and NaAc you need are 0.122M

Explanation:

pKa of acetic acid is 4.75, that means when amount of sodium acetate and acetic acid is the same, pH will be 4.75

Thus, you know [NaAc]i = [HAc]i

Now, using H-H equation, when pH = 3.75:

3.75 = 4.75 + log [NaAc] / [HAc]

0.1 = [NaAc] / [HAc]

10 [NaAc] = [HAc]

Thus, after the reaction  [HAc] must be ten times,  [NaAc].

Based in the reaction of NaAc with HCl

NaAc + HCl → HAc + NaCl

Moles of HCl added are:

1mL = 0.001L * (10mol /L) = 0.01 moles HCl.

That means moles of both compounds after the reaction are:

[NaAc] = [NaAc]i - 0.01 mol

[HAc] = [HAc]i + 0.01

Replacing these equations with the information you know:

[NaAc] = [NaAc]i - 0.01 mol

10[NaAc] = [NaAc]i + 0.01

Subtracting both equations:

9[NaAc] = 0.02mol

[NaAc] = 0.0022 moles.

Replacing in [NaAc] = [NaAc]i - 0.01 mol

0.0022mol = [NaAc]i - 0.01 mol

0.0122mol = [NaAc]i = [HAc]i

These moles in 100.00mL = 0.1000L:

[NaAc]i = [HAc]i = 0.0122mol / 0.100L =

0.122M

Thus, concentrations of HAc and NaAc you need are 0.122M

To create this buffer, you need to pipette 12.2mL of both 1.0M NaAc and 1.0M HAc and dilute this mixture to 100.0mL