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When performing titration between NaOH and H2SO4, both Na2SO4 and NaHSO4 are formed (both are soluble in water), is there any method to obtain crystals of NaHSO4? (Maybe by separating two soluble salts or varying the number of moles of NaOH?)

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  • $\begingroup$ It seems the question follows 2 different goals, the titration analytics and salt preparation. Can you clarify the question? $\endgroup$ – Poutnik Oct 10 '19 at 12:31
  • $\begingroup$ The question states that when titrating H2SO4 against NaOH, both Na2SO4 (H2SO4+ 2NaOH—> Na2SO4 + 2H2O) and NaHSO4 (H2SO4+ NaOH —> NaHSO4 + H2O) will form. Outline a method to obtain crystals of NaHSO4. $\endgroup$ – Raymand Tey Oct 10 '19 at 13:27
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It seems the most used way is reaction of sulphuric acid with salts of volatile mineral acids :

$$\ce{H2SO4 + NaCl -> NaHSO4 + HCl ^}$$

Reaction with excess of the acid

$$\ce{H2SO4 + NaOH ->[H2SO4] NaHSO4 +H2O}$$

has several drawbacks:

  • It is much more exothermic than the former one

  • It releases extra water

  • It needs somewhat diluted solution not to be violent.

  • It needs evaporation of water from quite concentrated

The interesting option could be recrystallization of $\ce{Na2SO4}$ (anhydrate) from $\ce{H2SO4}$.

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  • $\begingroup$ Thank you so much! Another question I want to ask would be is there a way to separate two soluble salts? Or in this case, Na2SO4 and NaHSO4? $\endgroup$ – Raymand Tey Oct 10 '19 at 15:04
  • $\begingroup$ $\mathrm{p}K_\mathrm{a2}$ of $\ce{H2SO4}$ is cca 2.0. There will be the minor concentration of $\ce{SO4^2-}$ in $\ce{H2SO4}$ so solubility product of $\ce{NaHSO4}$ should be reached much sooner. $\endgroup$ – Poutnik Oct 10 '19 at 15:09

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