# Concentration of weak acid neutralized by strong base [closed]

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?

• 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/… May 15, 2021 at 21:53
• A strong base would be something like sodium hydroxide which which would react to neutralize both hydrogens of the $\ce{H2SO3}$ molecule.
– MaxW
May 15, 2021 at 21:59
• @MaxW so how should I change my process based on that information? Something like [H+]^2 [A-] / [HA] = Ka? May 15, 2021 at 22:16
• Forget Ka. Just divide 0.186 by two. May 15, 2021 at 22:49
• Assuming NaOH for the base, the overall chemical reaction would be: $$\ce{H2SO3 + 2NaOH -> Na2SO3 + 2H2O}$$
– MaxW
May 16, 2021 at 17:59

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