Question

A $20.0ml$ sample of a $0.20M$ solution of the weak diprotic acid $H_{2}A$ is titrated with $0.250MNaOH$. The solution of the second equivalent point is:

• A. $0.10MNaHA$
• B. $0.153MN{a}_{2}A$
• C. $0.10MN{a}_{2}A$
• D. $0.0769MN{a}_{2}A$

Answer 1

Answer: (D)

$0.0769MN{a}_{2}A$

For $2^{nd}$ equivalent point, the reaction is:

$H_{2}A+2NaOH\to N{a}_{2}A+H_{2}O$

From the above reaction-

$2\times$ no. of milimoles of $H_{2}A=$ no. of milimoles of $NaOH$

$2n_{\left(H_{2}A\right)}=n_{\left(NaOH\right)}$

$2\times 20ml\times 0.2M=0.25M\times Vml$

$V=32ml$

$n_{\left(N{a}_{2}A\right)}=n_{\left(H_{2}A\right)}$

$\Rightarrow n_{\left(N{a}_{2}A\right)}=4milimole=20ml\times 0.2$

Concentration of $N{a}_{2}A$,

$\left[N{a}_{2}A\right]=\frac{milimoles}{V\left(inml\right)}$

$=\frac{4}{(20+32)}=0.0769M$

$\left[N{a}_{2}A\right]=0.0769M$