# Neutralisation of weak acid by base

I'm still working on my concepts related to chemical equilibrium. I came across this question:

10mL of a weak acid HA is 20% dissociated in water. This solution is completely neutralised by 10 mL of $$10^{-3}$$ M of NaOH.

Is [$$\ce{HA}$$] = $$10^{-3}$$ M or is [$$\ce{H^+}$$] =$$10^{-3}$$?

Why is it wrong to say that since NaOH completely dissociates, and solution is neutral, [$$\ce{OH^-}$$] =$$10^{-3}$$ M= [$$\ce{H^+}$$]?

Here, [.] denotes concentration.

• If your weak acid is neutralized by $\ce{NaOH}$, the result does not depend on the percentage of dissociation before neutralization. Here your initial acid is $\pu{10^{-3} M}$, as you stated. But the initial concentration of $\ce{H+}$ is $20$% of this value, so : $\ce{[H+] = 2·10^{-4}} M$ Mar 22 '21 at 11:09
• Neutralised is ambiguous here. In aqueous acid/base solutions neutralized usually means that the pH was adjusted to 7.0. It could also mean that as many moles of base were added as there were moles of acid. In a titration this is referred to as the equivalence point. The final pH will be greater than 7 due to the reaction $$\ce{A- + H2O <=> HA + OH-}$$
– MaxW
Mar 22 '21 at 17:49
• @MaxW you're correct; here the equivalence point is where the equivalents of the acid and base will be equal. The hydrolysis will yield a solution whose ph is $$ph =7+\frac{1}{2}\biggr[pka + log[HA]\biggr]$$ Mar 22 '21 at 19:54
• @MaxW This is why I got confused. Now I understood it. Thanks a lot!
– sam
Mar 23 '21 at 3:53

As @Maurice pointed out, the result of a neutralization reaction is not impacted by the dissociation percentage.

Is $$\pu{[}\ce{HA}{] = 10^{−3} M}$$ or is $$\pu{[}\ce{H+}{] = 10^{−3}}$$?

$$\pu{[}\ce{HA}{] = 10^{-3}}$$.

Always remember, neutralization in aqueous medium depends upon the number of equivalents of acid and base: $$N_1V_1=N_2V_2$$ Since $$V_1=V_2$$, therefore, $$\pu{[}\ce{HA}{]=[}\ce{NaOH}{]}$$.

Why is it wrong to say that since $$\ce{NaOH}$$ completely dissociates, and solution is neutral, $$[\ce{OH−}]$$ $$\pu{= 10^−3 M = [}\ce{H+}]$$?

The secret lies in the name of the concept itself. Chemical equilibrium. To attain an equilibrium state, all favorable chemical reactions are first carried out. The remaining products/ions constitute the equilibrium state. Coming back to the question, the neutralisation reaction between $$\ce{HA}$$ and $$\ce{NaOH}$$ is carried out first:

$$\ce{HA + NaOH ->NaA +H2O}$$ The resultant ions constitute the equilibrium, which, are none unless you consider the hydrolysis of the resultant salt of a weak acid/strong base.

Since your weak acid is neutralized to the end point by a strong base, $$\ce{NaOH}$$, the final reding does not depend on the percentage of dissociation of the weak acid before neutralization. Suppose your ionization of the weak acid is as follows:

$$\ce{HA + H2O <=> H3O+ + A-} \tag1$$

Thus at the end of neutralization:

$$\text{amount of \ce{NaOH} needed} = \text{amount of \ce{HA} at equilibrium} + \text{amount of \ce{H3O+} at equilibrium}$$

Since volumes of both solutions are $$\pu{10 mL}$$ each:

$$[\ce{HA}] + [\ce{H3O+}] = [\ce{NaOH}] = \pu{10^{-3} M}$$

Here, $$[\ce{HA}]$$ and $$[\ce{H3O+}]$$ concentration of each at the equilibrium. My argument is when you get $$\pu{10 mL}$$ of aqueous solution of a weak acid $$\ce{HA}$$, all three species of equation $$(1)$$ is in equilibrium, regardless of its ionization percentage. Therefore, sum of $$[\ce{HA}]$$ and $$[\ce{H3O+}]$$ is equal to $$\pu{10^{-3} M}$$, which is $$[\ce{OH-}]$$ since $$\ce{NaOH}$$ has completely dissociated in an aqueous solution.

Note that, the percentage is given to calculate the $$K_\mathrm{a}$$ of the acid. In this case (your highlighted question is not completed), $$K_\mathrm{a} = \frac{0.2 \times 0.2}{1-0.2} = 0.05$$