1. Tardigrade
  2. Question
  3. Chemistry
  4. 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 MBa ( OH )2 ?

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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 $0.10 \,MBa ( OH )_{2} ?$

Solution:

Millmoles of $H^{+}$produced $=20 \times 0.05=1$
Millmoles of $O H^{-}$produced $=30 \times 0.1 \times 2=6$
(Each $Ba ( OH )_{2}$ gives $2 OH ^{-}$).
Millmoles of $O H^-$ remaining in solution $=6-1=5$
Total volume of solution $=20+30=50\, mL$
$\left[O H^{-}\right]=\frac{5}{50}=0.1\, M$