Homework Statement

To find pH at the equivalence point in a titration it seems there are two different ways to do the calculation depending on if we use one proton transfer equation (that of the conjugate acid) or another (that of the original base analyte).

Relevant Equations

I would like to figure out why it is that the latter way requires more steps whereas the previous way is simpler.

Methylamine has a basicity constant of ##K_b=5.6\times 10^{-4}##.

We start with a 75mL sample of 0.500M solution of methylamine in water.

We titrate with a 0.205 solution of HCl.

The pH of the initial sample is easily calculated from the expression

$$\mathrm{K_b=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}=\frac{x^2}{0.5-x}}$$

$$\implies x=1.65\times 10^{-2}$$

$$\implies \mathrm{pH=12.22}$$

Now, consider the task of finding the pH at the equivalence point.

The 75mL sample contains 0.0375mol of methylamine which must be completely neutralized by added HCl.

The volume of titrant required comes out to 0.183L.

The neutralization of the analyte generates 0.0375mol of its conjugate acid, ##\mathrm{CH_3NH_3^+}##.

At this point, I have questions.

There are two ways to move forward.

Method 1

One way is to use the proton transfer equation for the conjugate acid.

$$\mathrm{CH_3NH_3^+(aq)+H_2O(l)\rightleftharpoons CH_3NH_2(aq)+H_3O^+(aq)}$$

with ##K_a=K_w/K_b##.

We start with a concentration of ##c_1=\frac{0.0375}{0.075+0.183}##M of the conjugate acid, and none of the products (this is an approximation since there is some non-zero hydronium concentration).

We solve

$$K_a=\frac{x^2}{c_1-x}$$

for ##x## and find that the pH is 5.79.

Ok, that was fairly simple and easy.

Method 2

What if instead of the proton transfer for the conjugate acid we use the proton transfer for methylamine to compute the pH at the equivalence point?

It seems that for some reason this way is more complicated. And I am trying to figure out the reason why it requires extra steps.

We have the equation

$$\mathrm{CH_3NH_2(aq)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq)+OH^-(aq)}$$

The naive approach here (and I want to know precisely why it is naive) is to solve

$$\mathrm{K_b=\frac{([CH_3NH_3^+]_{init}-x)([OH^-]_{init}-x)}{x}}$$

where

$$\mathrm{[CH_3NH_3^+]_{init}=\frac{0.0375mol}{(0.075+0.183)L}}$$

$$\mathrm{[OH^-]_{init}=\frac{1.65\times 10^{-2}mol}{(0.075+0.183)L}}$$

and when we solve for ##x## we find

$$\mathrm{x=4.79\times 10^{-3}}$$

$$\implies \mathrm{[OH^-]=5.93\times 10^{-7}}$$

which gives a pH=7.77.

Is this just plain completely wrong or is it just very, very imprecise?

If we want to get it right, we need to use more equations. Here they are
$$\mathrm{K_b=5.6\times 10^{-4}=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}}$$

$$\mathrm{K_2=[H_3O^+][OH^-]}$$

$$\mathrm{[CH_3NH_3^+]_{init}=[CH_3NH_3^+]+[H_3O^+]-[OH^-]}$$

$$\mathrm{[CH_3NH_3^+]=[CH_3NH_3^+]+[CH_3NH_2]}$$

The latter two are charge balance and material balance.

We end up with the expression

$$\mathrm{K_b=\frac{([CH_3NH_3^+]_{init}+[OH^-]-[H_3O^+])[OH^-])}{[CH_3NH_3^+]_{init}-([CH_3NH_3^+]_{init}+[OH^-]-[H_3O^+])}}$$

which we can solve for ##\mathrm{[H_3O^+]}## using ##\mathrm{[OH^-]=\frac{K_w}{[H_3O^+]}##.

These equations give us the 5.79 of pH.

So why exactly do we have to go through all the extra steps in method 2 but not in method 1?

$$\mathrm{K_a=\frac{K_w}{K_b}=\frac{[H_3O^+][CH_3NH_2]}{[CH_3NH_3^+]}=\frac{x^2}{[CH_3NH_3^+]_{init}-x}}\tag{1}$$

$$\mathrm{K_b=\frac{([CH3NH_3^+]_{init}-x)K_w}{x^2}}\tag{1a}$$

to the one solved in method 2

$$\mathrm{K_b=\frac{([CH_3NH_3^+]_{init}+[OH^-]-[H_3O^+])[OH^-])}{[CH_3NH_3^+]_{init}-([CH_3NH_3^+]_{init}+[OH^-]-[H_3O^+])}}\tag{2}$$

$$\mathrm{=\frac{([CH_3NH_3^+]_{init}+\frac{K_w}{x}-x)\frac{K_w}{x})}{[CH_3NH_3^+]_{init}-([CH_3NH_3^+]_{init}+\frac{K_w}{x}-x)}}\tag{3}$$

$$\mathrm{=\frac{([CH_3NH_3^+]_{init}+\frac{K_w}{x}-x)\frac{K_w}{x}}{-\frac{K_w}{x}+x}}\tag{4}$$

Suppose ##\mathrm{[OH^-]=\frac{K_w}{x}\approx 0}##. Then (4) seems to become (1a).

Therefore, method 2 takes into account something that method 1 does not but this something is irrelevant for the final result.

We can also compare to the equation solved in the "naive" approach to method 2

$$\mathrm{K_b=\frac{([CH_3NH_3^+]_{init}-x)([OH^-]_{init}-x)}{x}}\tag{5}$$

I dunno. I give up for now.

It is a salt, so the hydrolysis occurs only to the conjugate acid.

BNH₃+ + H₂O <==> BNH₂ + H₃O+

The equilibrium is given by the constant Kh = [BNH2][H+]/[BNH₃+]
This constant is unknown, but as:

BNH₂ + H₂O <==> BNH3+ + OH

kb = [BNH₃+][OH-]/[BNH₂]

and

kw = [H+][OH-]

Then kh = kw/kb which in this case = 1 x 10^-14/5.6 x 10^-4 = 1.79 x 10^-11

Hence:

[BNH₂][H+]/[BNH₃+] = 1.79 x 10^-11

If we then make the assumption that the concentration of the salt conjugate acid ion is stoichiometric (as it is ionic) then we can determine its concentration from the volume and concentration of acid and base mixed together initially.

75mL sample of 0.500M solution of methylamine is neutralised by 0.183 dm³ of HCl
Hence total volume = 0.258 dm³

And the total mol of salt formed = 0.0375 mol

So the concentration of the salt = 0.145 mol dm^-3

Substituting into the Kh equation:

[BNH₂][H+]/[BNH₃+] = [H+]² = 1.79 x 10^-11 x 0.145 = 2.60 x 10^-12

Hence [H+] = 1.61 x 10^-6

pH = 5.79

Which is what you would expect from a neutralisation of a strong acid with a weak base. (pH less than 7)
(apologies for poor formatting)

I don't know if it fully answers the question, as I haven't been through it all, but

is wrong. The concentration of OH^- in the starting solution is 1.65 x 10^-2 M. The amount is not 1.65 x 10^-2 mol.
[OH^-]_init = 1.65 x 10^-2 M x 0.075/(0.075 + 0.183)

We add 183mL of titrant to reach the equivalence point.

This titrant neutralizes the entirety of the methylamine present (0.0375 mol) and generates 0.0375 mol of its conjugate acid, ##\mathrm{CH_3CO_2^-}##.

In order to calculate the resulting equilibrium using the version of method 2 in the OP that gave the incorrect result seems to include some initial concentration for hydroxide.

Indeed, ##\mathrm{[OH^-]=1.65\times 10^{-2}M}## is not the correct concentration after the titration.

Rather, ##\mathrm{[OH^-]=1.65\times 10^{-2}\times \frac{0.075}{0.075\times 0.183}}## seems like more correct.

However, this also does not generate the correct pH, which comes out to 9.27.

What initial concentration of hydroxide would you use?