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I would really appreciate some help with how to approach this:

I'm given that:

Methylamine (CH3NH2) is a weak base used in the manufacture of pesticides. Taking into account that the degree of dissociation of said compound is 21%, calculate:

Kb=6.1*10^-4 (This is likely an arbitrary number assigned to each student, rather than being the correct Kb, as is the degree of dissociation).

a) The initial concentration of methylamine necessary to prepare 175 mL of solution. - This I have managed to solve, my answer is 0.01093M, the autocorrect mechanism says it's right

Edit - Here is what I've done:

CH3NH2(aq) | + | H2O | <> | OH- | + | CH3NH3+(aq)

 1      |              -   |                 0   |       |         0     |
 1-α    |              -   |                 +α  |       |         +α    |
Co(1-α) |              -   |                 Coα |       |       Coα     |

Kb= 6x10^-4 = (Co*.21)^2/Co(1-.21)

Solving this quadratic equation yields Co = 0.01093M; Which is correct according to the autocorrect mechanism.

b)How many moles of methylamine chloride (CH3NH3Cl) must be added to the previous solution to obtain 217 mL of final solution with a pH = 11.16? - This is what I cannot fathom, I've tried making ICE charts and the Henderson-Hasselbalch equation, though I'm not having any luck, it's likely that I am doing it wrong though

Edit 2

This I've tried many times to do and feel like I'm going round in circles, hence this post.

I need to create from the previous solution above 217ml of a new buffer solution with Ph 11.16.

Therefore:

POH = 2.84 so;

10^-2.84 = 0.001445439

Which should be the OH- in the new solution.

POH = Pkb + log [conjugate acid / weak base] Kb = 6.1x10^-4 PKb = -logKb = 3.21467

-0.37 = Log[CA] - Log [0.01093]

10^-2.34 = 0.004612644 M

Since I need 217ml, I multiply; 0.004612644 by; .217 giving; 0.00100943 moles of conjugate acid. This is wrong according to the autocorrect.

I have tried doing an ICE chart again , which I won't rewrite here, and I obtained a similar figure. That said, I have also obtained other figures which are also wrong. So I think there must be some flaw in my reasoning here, which I haven't been able to identify myself.

Edit 3 - I won't comment on this part since, it flows from the previous part. Any comments will be hugely appreciated!

If 2.71 mL of 0.351 M HNO3 are added to the solution in part b), what will the pH of the final solution be? This I feel more confident about, but cannot calculate it without first getting b)

Thanks in advance!

Edit 4 - Base at equilibrium:

****CH3NH2(aq) + H2O <> OH- + CH3NH3+(aq)****
0.01093 - 0 0
-.0022953 - 0.01093 x.21 0.01093 x.21
0.0086347 - 0.0022953 0.0022953

Kb=6.1x10^-4

Desired OH- = 0.001445439

Kb = [OH-][BH+]/[CH3NH2]

BH+ = 6.1x10^-4(0.0086347/0.001445439) = 0.003643991M

0.003643991M x .217 = .000790746 moles needed (wrong according to autocorrect)

Likewise

2.84 = Pkb + Log(conjugate acid/weak base)

[CA] = (0.0086347/10^.374670165) = 0.003643989M

0.003643989M x.217 = 0.000790745 moles (basically the same number)

There must be something else I am getting wrong, could it be to do with the volume, changing from .175L or original solution to .217L of new solution

Edit 5

I finally got it, the problem was I was using concentrations for the calculations, and since the volume changes, I should have been using moles.

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    $\begingroup$ Practical advice: Approach your task by algebraic solution first and use literal values in the end, when all the formulas are ready. It has several benefits. $\endgroup$ – Poutnik Jan 4 at 12:41
  • $\begingroup$ You have started from the wrong end. Start with calculation of the final solution concentration of the base. The from pKb, pOH and [B] you compute [BH+] and from it the molar amount. $\endgroup$ – Poutnik Jan 4 at 15:19
  • $\begingroup$ Thanks, do I need to calculate the degree of dissociation again for the new solution also? Or should I use the 21% given? $\endgroup$ – Theo Clarke Jan 4 at 15:34
  • $\begingroup$ @21% is a particular value for the very particular case. Start with equations of mass and charge conservation and the equlibrium equations. Then involve affordable simplifications. $\endgroup$ – Poutnik Jan 4 at 15:43
  • $\begingroup$ Thanks, I hope I understood you correctly, I have tried to do as you say, though I could easily have made a mistake. Unfortunately I'm still unable to achieve the correct figure, I have added it above. $\endgroup$ – Theo Clarke Jan 4 at 16:54

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