TPR G.Chem Question.

[theA1doctor]

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I am not getting this one.

A 2 M solution of the potent organic pollutant sodium methylthiolate (NaSCH3) is titrated to its endpoint with 2 M HCl. Which of the following best approximates the pH of the resultant solution? (pKb of SCH3– = 3.6)

Correct Answer: –log (1 x 10–5.2)

Why is it 1M of methylthiol?

Here is the solution.

The result of the titration will be a 1 M solution of methylthiol, HSCH3 since the initial 2 M solution has been doubled in volume during the titration. Since we know the pKb of its conjugate base SCH3– is 3.6, then we know the pKa of HSCH3 is 10.4. Therefore, at equilibrium:

10–10.4 = (x)(x)/(1–x)

Where x = [H+] = [SCH3–], at equilibrium. As the resultant value of x is negligibly small compared to 1, we ignore x in the denominator and say that x = [H+] = [SCH3–] = (10–10.4)(0.5) giving 1 x 10–5.2. The answer –log (2 x 10–5.2) would be correct if the resultant solution was 2 M in concentration. The answer –log (1 x 10–1.8) is the result of using the pKb of the methylthiolate anion in the calculation rather than the pKa of HSCH3, and –log (1 x 10–3.6) would be the result of a similar a calculation using this pKb value, while not taking the square root.

 

[circulus vitios]

Sodium methylthiolate is a monoprotic base. HCl is a monoprotic acid.

Let's say we started with 26.3 mL of 2M sodium methylthiolate, that's 52.6 mmol of base.
We add 52.6 mmol of a monoprotic 2M acid (HCl) to reach the equivalence point.
52.6 mmol / 2M = 26.3 mL of HCl.

What's the total volume? 26.3 mL sodium methylthiolate + 26.3 mL HCl = 52.6 mL solution.
How much of the conjugate acid (methylthiol) do we have? 52.6 mmol.
What's the concentration of the conjugate acid? 52.6 mmol / 52.6 mL = 1M.

edit:

I have no idea wtf TPR is talking about with this (x)(x)/(1–x) nonsense. TBR's approach is the best: pH = pKA/2 - log([acid])/2.

We know that pKA + pKB = 14 for a conjugate acid and conjugate base pair. pKA is therefore 10.4.
We just found that the concentration of the conjugate acid is 1 M.

Plug and chug! pH = (10.4/2) - log(1)/2 = (10.4/2) = 5.2.

 

[theA1doctor]

Sodium methylthiolate is a monoprotic base. HCl is a monoprotic acid.

Let's say we started with 26.3 mL of 2M sodium methylthiolate, that's 52.6 mmol of base.
We add 52.6 mmol of a monoprotic 2M acid (HCl) to reach the equivalence point.
52.6 mmol / 2M = 26.3 mL of HCl.

What's the total volume? 26.3 mL sodium methylthiolate + 26.3 mL HCl = 52.6 mL solution.
How much of the conjugate acid (methylthiol) do we have? 52.6 mmol.
What's the concentration of the conjugate acid? 52.6 mmol / 52.6 mL = 1M.

edit:

I have no idea wtf TPR is talking about with this (x)(x)/(1–x) nonsense. TBR's approach is the best: pH = pKA/2 - log([acid])/2.

We know that pKA + pKB = 14 for a conjugate acid and conjugate base pair. pKA is therefore 10.4.
We just found that the concentration of the conjugate acid is 1 M.

Plug and chug! pH = (10.4/2) - log(1)/2 = (10.4/2) = 5.2.

OK, big thanks! I get it.

With the second part, TPR is setting up an acid disassociation equilibrium