I've got a question that should be simple enough but I must be doing something very wrong.

A $\pu{10mL}$ sample of $\ce{H2SO3}$ is neutralized by $\pu{18.6mL}$ of a $\pu{0.10M}$ strong base. Find the concentration of the acid.

My solution:

  1. Given the information, I calculate $\pu{1.86E-3 mol}$ of $\ce{OH-}$ used for this reaction, so an equivalent amount of $\ce{H+}$ ions is assumed. Therefore the $\ce{[H+]} = 1.86 \times 10^{-3} / \pu{0.01L} = \pu{0.186M}$.

  2. $K_\mathrm{a}$ of sulphurous acid is $1.5\times 10^{-2}$. $K_\mathrm{a} = \frac{\ce{[H+][A-]}}{ \ce{[HA]}}$. Assuming that $\ce{[H+]} = \ce{[A-]}$, the $\ce{[H2SO3]}$ at end of reaction is $\pu{2.3064M}$.

However, this is obviously wrong compared to the given answer, $\pu{0.093M}$. Is the "concentration of acid" not $\ce{[HA]}$? Am I making any incorrect assumptions?

  • 3
    $\begingroup$ Your title is very wrong. Short of that, you have a common misconception that [H+] matters in the neutralization business. It doesn't. Also, see chemistry.stackexchange.com/questions/60407/… $\endgroup$ Commented May 15, 2021 at 21:53
  • $\begingroup$ A strong base would be something like sodium hydroxide which which would react to neutralize both hydrogens of the $\ce{H2SO3}$ molecule. $\endgroup$
    – MaxW
    Commented May 15, 2021 at 21:59
  • $\begingroup$ @MaxW so how should I change my process based on that information? Something like [H+]^2 [A-] / [HA] = Ka? $\endgroup$ Commented May 15, 2021 at 22:16
  • 4
    $\begingroup$ Forget Ka. Just divide 0.186 by two. $\endgroup$ Commented May 15, 2021 at 22:49
  • $\begingroup$ Assuming NaOH for the base, the overall chemical reaction would be: $$\ce{H2SO3 + 2NaOH -> Na2SO3 + 2H2O}$$ $\endgroup$
    – MaxW
    Commented May 16, 2021 at 17:59

1 Answer 1


A one loose fault of the question is not giving the nature of strong acid. Thus, it is safe to assume that $\ce{NaOH}$ is the strong base, and the overall chemical reaction can be written as:

$$\ce{H2SO3 + 2 NaOH -> Na2SO3 + 2 H2O}$$

Thus, each $\pu{mol}$ of $\ce{H2SO3}$ need $\pu{2 mol}$ of $\ce{NaOH}$ to neutralize. You have correctly calculate $\pu{1.86 \times 10^{−3} mol}$ of $\ce{OH−}$ used for this neutralization reaction. Therefore, amount of $\ce{H2SO3}$ presence in the $\pu{10.0 mL}$ of solution is $\frac12 \times \pu{1.86 \times 10^{−3} mol} = \pu{9.30 \times 10^{−4} mol}$. Thus, molarity of $\ce{H2SO3}$ solution is:

$$\frac{\pu{9.30 \times 10^{−4} mol}}{\pu{10.0 \times 10^{−3} L}} = \pu{9.30 \times 10^{−2} mol L-1} = \pu{0.093 M}$$


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