Question

If $50{\text{ ml}}$ of $0.1{\text{ M NaCl}}$ and $50{\text{ ml}}$ of $0.1{\text{ M
BaC}}{{\text{l}}_{\text{2}}}$ are mixed, molarity of chloride ion in the resulting solution will be:
A. 0.2M
B. 0.3M
C. 0.15M
D. 0.1M

Answer 1

Hint: Use the molar concentration and volume of each solution and calculate the moles of each ${\text{NaCl}}$ and ${\text{BaC}}{{\text{l}}_{\text{2}}}$. Using a stoichiometric ratio calculate the moles of chloride ions present in each solution. From the total moles of chloride ion and total volume of solution calculate the molar concentration chloride ion.
We can use the Formula:
${\text{Molarity = }}\dfrac{{{\text{moles of solute}}}}{{{\text{L of solution }}}}$

Step by step answer:
Calculate the moles of ${\text{NaCl}}$ as follows:
${\text{Molarity = }}\dfrac{{{\text{moles of solute}}}}{{{\text{L of solution }}}}$
We have given $50{\text{ ml}}$ of$0.1{\text{ M NaCl}}$.So, convert the volume of ${\text{NaCl}}$solution from ml to L.
1L = 1000ml
So, $50{\text{ ml }} \times \dfrac{{1{\text{ L}}}}{{1000{\text{ ml }}}} = 0.05{\text{ L}}$
Now, substitute $0.1{\text{ M NaCl}}$ for molarity and $0.05{\text{ L}}$ for a volume of solution and calculate moles of ${\text{NaCl}}$.
\[{\text{moles of NaCl = molarity }} \times {\text{ L of solution }}\]
\[{\text{moles of NaCl = 0}}{\text{.1M }} \times {\text{ 0}}{\text{.05L }}\]
\[{\text{moles of NaCl = 0}}{\text{.005 mol}}\]
Now, using the stoichiometric ratio of${\text{NaCl}}$to \[{\text{C}}{{\text{l}}^{\text{ - }}}\] calculate the moles of chloride ions.
\[1{\text{ mol NaCl = 1 mol C}}{{\text{l}}^{\text{ - }}}\]
So, \[{\text{0}}{\text{.005 mol NaCl = 0}}{\text{.005 mol C}}{{\text{l}}^{\text{ - }}}\]
Similarly, calculate the moles of ${\text{BaC}}{{\text{l}}_{\text{2}}}$ as follows:
We have given $50{\text{ ml}}$ of$0.1{\text{ M BaC}}{{\text{l}}_{\text{2}}}$.So convert the volume of ${\text{NaCl}}$solution from ml to L.
1L = 1000ml
So, $50{\text{ ml }} \times \dfrac{{1{\text{ L}}}}{{1000{\text{ ml }}}} = 0.05{\text{ L}}$
Now, substitute $0.1{\text{ M BaC}}{{\text{l}}_{\text{2}}}$ for molarity and $0.05{\text{ L}}$for a volume of solution and calculate moles of ${\text{BaC}}{{\text{l}}_{\text{2}}}$.
\[{\text{moles of BaC}}{{\text{l}}_{\text{2}}}{\text{ = molarity }} \times {\text{ L of solution }}\]
\[{\text{moles of BaC}}{{\text{l}}_{\text{2}}}{\text{ = 0}}{\text{.1M }} \times {\text{ 0}}{\text{.05L }}\]
\[{\text{moles of BaC}}{{\text{l}}_{\text{2}}}{\text{ = 0}}{\text{.005 mol}}\]
Now, using the stoichiometric ratio of${\text{BaC}}{{\text{l}}_{\text{2}}}$to \[{\text{C}}{{\text{l}}^{\text{ - }}}\] calculate the moles of chloride ions.
\[1{\text{ mol BaC}}{{\text{l}}_{\text{2}}}{\text{ = 2 mol C}}{{\text{l}}^{\text{ - }}}\]
So, \[{\text{0}}{\text{.005 mol BaC}}{{\text{l}}_{\text{2}}}{\text{ }} \times {\text{ }}\dfrac{{{\text{2 mol C}}{{\text{l}}^{\text{ - }}}}}{{1{\text{ mol of BaC}}{{\text{l}}_{\text{2}}}{\text{ }}}} = {\text{ 0}}{\text{.01 mol of C}}{{\text{l}}^{\text{ - }}}\]
Calculate the molar concentration of chloride ion as follows:
\[{\text{total moles of C}}{{\text{l}}^{\text{ - }}} = {\text{ 0}}{\text{.005 mol + 0}}{\text{.01 = 0}}{\text{.015 mol C}}{{\text{l}}^{\text{ - }}}\]

\[{\text{total volume of solution }} = {\text{ 0}}{\text{.05L + 0}}{\text{.05L = 0}}{\text{.1 L}}\]

${\text{[C}}{{\text{l}}^{\text{ - }}}{\text{] = }}\dfrac{{{\text{0}}{\text{.015 mol C}}{{\text{l}}^{\text{ - }}}}}{{0.1{\text{ L}}}}$

${\text{[C}}{{\text{l}}^{\text{ - }}}{\text{] = 0}}{\text{.15 M}}$

So, the molarity of chloride in the mixture of $50{\text{ ml}}$ of $0.1{\text{ M NaCl}}$ and $50{\text{ ml}}$ of $0.1{\text{ M BaC}}{{\text{l}}_{\text{2}}}$ is ${\text{0}}{\text{.15 M}}$.

Thus, the correct option is (C) ${\text{0}}{\text{.15 M}}$

Note: Concentration of solution is the amount of solute present in a given amount of solution. The concentration of the solution is expressed in various units; there are various formulas to calculate the concentration. It is very important to use the correct formula of concentration and correct values of the amount of solute and amount of solution with correct units.