Question

What volume of 0.02 M $Ba(OH{)}_2$ should be mixed with 40 ml of 0.02 M $NaOH$ and $20$ ml of 0.01 M $HN{O}_3$ to get solution that contain $pH=12.3$?

[Use : log 2 = 0.3]

• A. 30 ml
• B. 40 ml
• C. 50 ml
• D. 60 ml

Answer 1

The correct option is A 30 ml

We know that $p^H+p^{OH}=14$

$p^{OH}=1.7$

$\left[O{H}^{-}\right]=$ Anti $\log \left(1.7\right)$

$=2\times {10}^{-2}$

So not $\left[O{H}^{-}\right]=2\times {10}^{-2}$

taking $100\%$ dissociation of $Ba(OH{)}_2$

Using mole conservation rule

$m_i{v}_i=m_f{v}_f$

$2\times {10}^{-2}\times 2\times v+2\times {10}^{-2}\times 40-20\times {10}^{-2}=2\times {10}^{-2}(60+v)$

$60+4v=120+2v$

$2v=60$

$v=30$