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Calculate the pH of a mixture of $75\:\mathrm{cm^3}$ of methanoic acid solution of concentration $0.10\:\mathrm{mol\:dm^{-3}}$ and $75\:\mathrm{cm^3}$ of sodium hydroxide solution of concentration $0.050\:\mathrm{mol\:dm^{-3}}$. The $\mathrm{p}K_\mathrm{a}$ of methanoic acid is $3.75$.

I got $0.00375\:\mathrm{mol}$ of $\ce{HCOONa}$, the salt, made. The concentration of that is $\dfrac{0.00375}{150/1000}=0.025\:\mathrm{mol\:dm^{-3}}$(?)

Using Henderson–Hasselbalch: $\mathrm{pH}=3.75+\log\dfrac{0.025}{0.10}=3.15$

The answer says $3.75$, is that a typo?

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    $\begingroup$ I don't know if it is a typo, but you do have an error in the Henderson Hasselbach equation. You forgot that the total concentration of methanoic acid changes when you add the sodium hydroxide. $\endgroup$
    – LDC3
    Commented Sep 21, 2014 at 14:39
  • $\begingroup$ Thanks, had a look at it and tried it again, got 3.45 as pH instead. Do you mean that the concentration changed because the methanoic acid was in excess, so the concentration would be 0.00375/(75/1000)? Because that's what I did. $\endgroup$
    – Jim
    Commented Sep 21, 2014 at 14:57
  • $\begingroup$ Yes, the methanoic acid is in excess. Since you have equal volumes of methanoic acid and sodium hydroxide, I took a shortcut to determine that you start with twice as much methanoic acid than sodium hydroxide. You still must have something wrong with the equation, since the calculation of $pK_a$ is when there is the same amount of acid as the salt. $\endgroup$
    – LDC3
    Commented Sep 21, 2014 at 15:33
  • $\begingroup$ Upon reading about the equation, I realized that it is not a typo (and you should get the value 3.75). So there is still an error in the equation. Determine what each variable represents in the HH equation. ncbi.nlm.nih.gov/pmc/articles/PMC3747999 $\endgroup$
    – LDC3
    Commented Sep 21, 2014 at 18:56
  • $\begingroup$ @Jim The only problem remaining now is that your concentration of methanoic acid is 0.00376/(150/1000). You forgot that its volume increases when the base is added. Lots of places to make small mistakes in these questions. $\endgroup$
    – Jason Patterson
    Commented Sep 29, 2014 at 16:53

1 Answer 1

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One will need the Henderson Hasselbach equation which is given as

$$\text{pH}=pK_a+\log\frac{\ce{A-}}{\ce{HA}}$$

where HA and A- are the concentrations of the conjugate acid and conjugate base.

First we find our moles of each.

$\mathrm{{mol_ {acid}}=75~cm^3\times(0.10~mol~dm^{-3})\times(10cm)^{-3}\ dm^{3} = 0.0075\ mol\ \times \dfrac{1000\ mmol}{1\ mol} = \ 750~mmol}$

same thing for base: $\mathrm {mol_{base} = 375} $ mmol

Set up an ICA (not ICE) to handle your limiting reagent: (Write down the chemical equation if this is hard to see. One should react acid with the best base available)

  • $\ce{HA}$: $\pu{750 mmol -375 mmol} = \pu{375 mmol}$
  • $\ce{OH-}$: $\pu{375 mmol - 375 mmol} = \pu{0}$
  • $\ce{H2O}$: --------------------
  • $\ce{A-}$: $\pu{0 + 375 mmol} = \pu{375 mmol}$

In the Henderson Hasselbach equation moles can be used instead of molarity because both conjugates are in the same volume so they cancel:

$$\mathrm{pH}=3.75+\log\dfrac{375\ \mathrm{mmol\ \ce{A-}}}{375\ \mathrm{mmol\ } \ce{HA}}$$

$\mathrm{pH}=3.75$

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  • $\begingroup$ What's an ICA table? I can't find that on the internet. $\endgroup$
    – Nerdatope
    Commented May 20, 2015 at 3:49
  • $\begingroup$ ICA table is just a way to organize limiting reagent stoichiometry. It is typically done in mol or mmol. It stands for initial, change, after. $\endgroup$
    – PhysicalChemist
    Commented May 20, 2015 at 3:52
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    $\begingroup$ I think you have miscalculated the no. of moles for acid and base. In both cases, the volume times strength will be divided by 1000 and not multiplied. $\endgroup$
    – chail10
    Commented Feb 28, 2018 at 15:52
  • $\begingroup$ $\mathrm{{mol_ {acid}}=75~cm^3\times(0.10~mol~dm^{-3})\times(10~cm)^{-3}~\pu{dm^3} =\pu{ 0.0075mol}}$ $\endgroup$
    – Adnan AL-Amleh
    Commented Nov 27, 2018 at 5:02
  • $\begingroup$ Thanks, good catch! I did not clarify mmol vs mol; the answer is updated. $\endgroup$
    – PhysicalChemist
    Commented Nov 28, 2018 at 15:43

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