Thank you for reporting, we will resolve it shortly
Q.
What is the $ [OH^-] $ in the final solution prepared by mixing $20.0\, mL$ of $0.050\, M\, HCl$ with $30.0\, mL$ of $0.10\, M Ba(OH)_2 $ ?
AIPMTAIPMT 2009Equilibrium
Solution:
Number of milliequivalents of $HCl$
$=20 \times 0.050 \times 1=1$
Number of milliequivalents of $Ba ( OH )_{2}$
$=2 \times 30 \times 0.10=6$
$\left[ OH ^{-}\right.$] of final solution
Milliequivalents of $Ba ( OH )_{2}$
$=\frac{-\text { milliequivalents of } HCl }{\text { Total volume }}$
$=\frac{6-1}{50}=0.1 \,M$