Q64 / 200 · NEET 2009 (Screening) · Chemistry

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

  • Correct answer
    A
    $0.10 \mathrm{M}$
  • B
    $0.40 \mathrm{M}$
  • C
    $0.0050 \mathrm{M}$
  • D
    $0.12 \mathrm{M}$

Solution

Number of milliequivalents of $\mathrm{HCl}$ $$ =20 \times 0.050 \times 1=1 $$ Number of milliequivalents of $\mathrm{Ba}(\mathrm{OH})_2$ $$ =2 \times 30 \times 0.10=6 $$ $\left[\mathrm{OH}^{-}\right]$of final solution $$ \begin{aligned} & \text { milliequivalents of } \mathrm{Ba}(\mathrm{OH})_2 \\ & =\frac{- \text { milliequivalents of } \mathrm{HCl}}{\text { total volume }} \\ & =\frac{6-1}{50}=0.1 \mathrm{M} \\ & \end{aligned} $$

Hence, the correct answer is option: 'A'.

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