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 $10 \mathrm{~mL}$ of $10^{-3} \mathrm{M}$ of $\mathrm{NaOH}$.
Is $[\mathrm{HA}]=10^{-3} \mathrm{M}$ or is $\left[\mathrm{H}^{+}\right]=10^{-3} ?$
Why is it wrong to say that since $\mathrm{NaOH}$ completely dissociates, and solution is neutral, $\left[\mathrm{OH}^{-}\right]=$ $10^{-3} \mathrm{M}=\left[\mathrm{H}^{+}\right] ?$
Here, [.] denotes concentration.
As @Maurice pointed out, the result of a neutralization reaction is not impacted by the dissociation percentage.
Is [HA]=10−3M[HA]=10−3M or is [H+]=10−3[HX+]=10−3?
[HA]=10−3[HA]=10−3.
Always remember, neutralization in aqueous medium depends upon the number of equivalents of acid and base:
N1V1=N2V2N1V1=N2V2
Since V1=V2V1=V2, therefore, [HA]=[NaOH][HA]=[NaOH].
Why is it wrong to say that since NaOHNaOH completely dissociates, and solution is neutral, [OH−][OHX−] =10−3 M=[H+]=10−3 M=[HX+]?
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 HAHA and NaOHNaOH is carried out first:
HA+NaOH⟶NaA+H2OHA+NaOH⟶NaA+HX2O
The resultant ions constitute the equilibrium, which, are none unless you consider the hydrolysis of the resultant salt of a weak acid/strong base.