Calculate the pOH of a solution that results from mixing 46 mL of 0.11

Question

Calculate the pOH of a solution that results from mixing 46 mL of 0.11 M trimethylamine ((CH₃)₃N) with 17 mL of 0.13 M (CH₃)₃NHCl. The K_b value for (CH₃)₃N is 6.5 x 10^-5

The answer is 3.83

Answer 1

We have a weak base, (CH₃)₃N and we have the conjugate acid (salt) of that base, (CH₃)₃NH⁺

This creates a BUFFER. And for such a buffer, we can use a form of the Henderson Hasselbalch equation...

pOH = pKb + log [salt]/[base]

The pKb is the negative log of the Kb = -log Kb = -log 6.5x10-5 = 4.19 = pKb

Now, let's find the [salt] and the [base]

Total volume = 46 ml + 17 ml = 63 ml

Final [salt]: (17 ml)(0.13 M) = (63 ml)(x M) and x = 0.0351 M

Final [base]: (46 ml)(0.11 M) = (63 ml)(x M) and x = 0.0803 M

pOH = 4.19 + log (0.0351/0.0803)

pOH = 4.19 + (-0.359)

pOH = 3.83