Question

You need to produce a buffer solution that has a pH of 4.87. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of sodium acetate will you need to add to this solution? Ka for acetic acid is 1.8×10−5.

Answer 1

13.34 mmol NaOAc needed ;)

Step-by-step explanation:

Answer 2

13.34 mmol NaOAc needed

Step-by-step explanation:

Use the Henderson-Hasselbalch Equation to determine the [OAc⁻]:[HOAc] ratio that will give a pH = 4.87. Then use that ratio to calculate the amount of NaOAc needed to make the buffer solution. ([pKa(HOAc) = -log(Ka) = -log(1.8 x 10⁻⁵) = 4.75)

pH = pKa + log([OAc⁻]/[HOAc])

4.87 = 4.75 + log([OAc⁻]/[HOAc])

log([OAc⁻]/[HOAc]) = 4.87 - 4.75 = 0.1253

[OAc⁻]/[HOAc] = 10⁰°¹²³⁵ = 1.334

That is, for a OAc⁻/HOAc buffer with a pH of 4.87 the [OAc⁻]:[HOAc] ratio must be 1.334:1.000.

Given that the system already contains 10 mmol HOAc one can determine the amount of OAc⁻ needed relative to the starting amount of HOAc as follows:

1.334/1.000 = (mmol NaOAc)/10mmol HOAc

=> mmol NaOAc = (10mmol HOAc)(1.334/1.000) = 13.34mmol NaOAc needed.

Test Calculation:

Assuming Molar concentrations, the following verifies the NaOAc quantity needed ...

HOAc => H⁺ + OAc⁻

C(Bfr) 0.0100M [H⁺] 0.01334M

Ka(HOAc) = [H⁺][OAc⁻]/[HOAc] = [H⁺](0.01334)/(0.0100M) = 1.8 x 10⁻⁵

[H⁺] = [(1.8 x 10⁻⁵)(0.0100)/(0.01334)]M = 1.349 x 10⁻⁵M

pH = -log[H⁺] - -log(1.349 x 10⁻⁵) = -(-4.87) = 4.87 => QED