Question

$100\text{mL}$ of $1\text{M}$ $NaCl$ solution is mixed with $100\text{mL}$ of $3\text{M}$ $HCl$ solution and $200\text{mL}$ of $1\text{M}$ $CaC{l}_2$ solution. Which of the following statements is true?

• A. The ratio of concentrations of cations to anions is $\frac{3}{4}$
• B. The ratio of cations to anions is 2
• C. $\text{Molarity of}C{l}^{-}=\frac{1}{2}M$
• D. $\text{Molarity of}C{l}^{-}=\frac{3}{2}M$

Answer 1

The correct option is A The ratio of concentrations of cations to anions is $\frac{3}{4}$

$\text{Molarity}=\frac{\text{number of moles of solute}}{\text{volume of solution in mL}}\times 1000$

$\text{Moles of}NaCl=\frac{1\times 100}{1000}=0.1\text{mol}$

$\text{Moles of}HCl=\frac{3\times 100}{1000}=0.3\text{mol}$

$\text{Moles of}CaC{l}_2=\frac{1\times 200}{1000}=0.2\text{mol}$

Moles of cations in the mixed solution $=\text{moles of}N{a}^{+}+H^{+}+C{a}^{2+}$
$\text{Total moles of cations}=0.1+0.3+0.2=0.6\text{mol}$

Similarly,
Moles of anions in the mixed solution
$=\text{moles of}C{l}^{-}\text{from}NaCl+C{l}^{-}\text{from}HCl+C{l}^{-}\text{from}CaC{l}_2$
$\text{Total moles of anions}=0.1+0.3+(2\times 0.2)=0.8\text{mol}$
Ratio of cations to anions $=\frac{0.6}{0.8}=\frac{3}{4}$
Molarity of $C{l}^{-}$= $\frac{0.8\times 1000}{400}=2\text{M}$