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What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M methylamine, $\ce{CH3NH2}$, with 25.00 mL of 0.10 M methylammonium chloride, $\ce{CH3NH3Cl}$? Assume that the volume of the solutions are additive and that Kb = 3.70 × 10-4 for methylamine.

1) First, I multiplied 0.025 L with 0.10 M and received 0.0025 moles. I then divided 0.0025 moles with 0.050 L (The total volume of the solution) and received 0.05 M, the molar concentration of the methylamine and its conjugate acid in this mixture.

2) I then created an ICE Chart (my reactant was the methylamine so that I would not need to find the Ka) and I solved for "x" (which would give me the concentration of $\ce{OH}$ ions at equilibrium). First, I solved for "x" but I assumed that is was insignificant. In the end I received 10.57 as the pH, which was the wrong answer. Then when I solved for "x" using the quadratic formula (assuming that "x" was significant), I got 0.0003646 as my "x".

3) When I took the negative log of 0.0003646, I got a pOH of 3.46. I subtracted this number by 14, and I still received a pH of 10.57

4) I even attempted to use the Henderson Hasselbalch equation and still received 10.57 because the ratio of the base and acid is 1, meaning that the log is 0.

I don't understand what I am doing incorrect. The answer is supposed to be 10.87, can someone please assist me?

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    $\begingroup$ Didn't you say the answer is 10.57 and you got it? Twice? $\endgroup$ – Andy Apr 13 '15 at 2:42
  • $\begingroup$ Sorry, that was a typo. The answer is 10.87. $\endgroup$ – Smith Apr 13 '15 at 10:04
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Your approach is correct and the result you obtain is right. Checked here.

As you correctly affirm, since the ratio of the base and acid is $1$, log will be zero and therefore $\mathrm{pH}=\mathrm{p}K_\text{a}$, where $\mathrm{p}K_\text{a} = 10.57$ (derived from the value of $K_\text{b}$ you provide).

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