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  4. The molar concentration of chloride ions in the resulting solution of 300 mL of 3.0 M NaCl and 200 mL of 4.0 M BaCl2 will be

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Q. The molar concentration of chloride ions in the resulting solution of $ 300\, mL $ of $ 3.0\,M\, NaCl $ and $ 200\,mL $ of $ 4.0 \,M \,BaCl_2 $ will be

MHT CETMHT CET 2012

Solution:

$\underset{3.0\, M}{NaCl} ⇌ Na^+ +\underset{3.0 \,M }{Cl^-}$

$\underset{4.0\,M}{BaCl_2}⇌ Ba^{2+} + \underset{2\times 4.0\,M}{2Cl^-} $

$\therefore $ Molar concentration of

$Cl ^{-} =\frac{M_{1} V_{1}+M_{2} V_{2}}{V_{1}+V_{2}} $

$=\frac{3.0 \times 300+2 \times 4.0 \times 200}{300+200} $

$=\frac{900+1600}{500} $

$=5.0\, M$