Question

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

• A. 0.10M
• B. 0.40M
• C. 0.0050M
• D. 0.12M

Answer 1

Answer: (A)

0.10M

Answer 2

$20\, mL$ of $0.50\, M\,HCl =20 \times 0.050\,m\, mol$
$=1.0\, m\, mol =1.0$meq .of $HCl$
$30\, mL$ of $0.10\, M\, Ba ( OH )_{2}$
$=30 \times 0.1\, m\, mol$
$=3 m\, mol =3 \times 2\, meq$
$=6\, meq Ba ( OH )_{2}$
$1\, meq$ of $HCl$ will neutralize 1 meq of $Ba ( OH )_{2}$
$Ba ( OH )_{2}$ left $=5\, meq$

Tatal volume $=50\, mL$
$Ba ( OH )_{2}$ conc. in final solution

$=\frac{5}{50} N =0.1\, N =0.05\, M$
$\left[ OH ^{-}\right]=2 \times 0.05\, M =0.10\, M$

Alternatively,

$Ba ( OH )_{2}+2 HCl \rightarrow BaCl _{2}+2 H _{2} O$
$2\, m\, mol$ of $HCl$ neutralize $1\, m$ mole of $Ba ( OH )_{2}$
$1\, m\, mol$ of $HCl$ neutralize $0.5\, m mol$ of $Ba ( OH )_{2}$
$Ba ( OH )_{2}$ left $=3-0.5\, m\, mol =2.5\, m\, mol$
$\left[ Ba ( OH )_{2}\right]=\frac{2.5}{50} M =0.05\, M$

or $[ OH ]^{-}=2 \times 0.05=0.1\, M$