Question

What is the $\left[O{H}^{-}\right]$ in the final solution prepared by mixing $20.0mL\text{of}0.050M$ $HCl$ with $30.0mL\text{of}0.10MBa(OH{)}_2$?

• A. 0.10 M
  
• B. 0.40 M
• C. 0.005 M
  
• D. 0.12 M

Answer 1

The correct option is A 0.10 M


Given, $V_{HCl}=20mL,V_{Ba(OH{)}_2}=30mL,\left[HCl\right]=0.050M,\left[Ba\right(OH{)}_2]=0.10M$
$mmole=N\times V_{\left(inmL\right)}$
$mmole=M\times \text{valence factor}\times V_{\left(inmL\right)}$
mmole of $HCl=20\times 0.05=1$
$\text{mmole of}Ba(OH{)}_2=30\times 0.1\times 2=6$
$\because Ba(OH{)}_2\rightleftharpoons B{a}^{2+}+2O{H}^{-}$
$\therefore$ mmol of $Ba(OH{)}_2\text{left in mixture}=6-1=5$
Total volume $=20+30=50mL$
$N_{Ba(OH{)}_2}or\left[O{H}^{-}\right]=\frac{5}{50}=0.1M$