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Question: If \(50{\text{ ml}}\) of \(0.1{\text{ M NaCl}}\) and \(50{\text{ ml}}\) of \(0.1{\text{ M BaC}}{{\...

If 50 ml50{\text{ ml}} of 0.1 M NaCl0.1{\text{ M NaCl}} and 50 ml50{\text{ ml}} of 0.1 M BaCl20.1{\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

Explanation

Solution

Use the molar concentration and volume of each solution and calculate the moles of each NaCl{\text{NaCl}} and BaCl2{\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:

Molarity = moles of soluteL of solution {\text{Molarity = }}\dfrac{{{\text{moles of solute}}}}{{{\text{L of solution }}}}

Step by step answer: Calculate the moles of NaCl{\text{NaCl}} as follows:

Molarity = moles of soluteL of solution {\text{Molarity = }}\dfrac{{{\text{moles of solute}}}}{{{\text{L of solution }}}}

We have given 50 ml50{\text{ ml}} of0.1 M NaCl0.1{\text{ M NaCl}}.So, convert the volume of NaCl{\text{NaCl}}solution from ml to L.

1L = 1000ml

So, 50 ml ×1 L1000 ml =0.05 L50{\text{ ml }} \times \dfrac{{1{\text{ L}}}}{{1000{\text{ ml }}}} = 0.05{\text{ L}}

Now, substitute 0.1 M NaCl0.1{\text{ M NaCl}} for molarity and 0.05 L0.05{\text{ L}}for a volume of solution and calculate moles of NaCl{\text{NaCl}}.

moles of NaCl = molarity × L of solution {\text{moles of NaCl = molarity }} \times {\text{ L of solution }}

moles of NaCl = 0.1M × 0.05L {\text{moles of NaCl = 0}}{\text{.1M }} \times {\text{ 0}}{\text{.05L }}

moles of NaCl = 0.005 mol{\text{moles of NaCl = 0}}{\text{.005 mol}}

Now, using the stoichiometric ratio ofNaCl{\text{NaCl}}to Cl - {\text{C}}{{\text{l}}^{\text{ - }}} calculate the moles of chloride ions.

1 mol NaCl = 1 mol Cl - 1{\text{ mol NaCl = 1 mol C}}{{\text{l}}^{\text{ - }}}

So, 0.005 mol NaCl = 0.005 mol Cl - {\text{0}}{\text{.005 mol NaCl = 0}}{\text{.005 mol C}}{{\text{l}}^{\text{ - }}}

Similarly, calculate the moles of BaCl2{\text{BaC}}{{\text{l}}_{\text{2}}} as follows:

We have given 50 ml50{\text{ ml}} of0.1 M BaCl20.1{\text{ M BaC}}{{\text{l}}_{\text{2}}}.So convert the volume of NaCl{\text{NaCl}}solution from ml to L.

1L = 1000ml

So, 50 ml ×1 L1000 ml =0.05 L50{\text{ ml }} \times \dfrac{{1{\text{ L}}}}{{1000{\text{ ml }}}} = 0.05{\text{ L}}

Now, substitute 0.1 M BaCl20.1{\text{ M BaC}}{{\text{l}}_{\text{2}}} for molarity and 0.05 L0.05{\text{ L}}for a volume of solution and calculate moles ofBaCl2{\text{BaC}}{{\text{l}}_{\text{2}}}.

moles of BaCl2 = molarity × L of solution {\text{moles of BaC}}{{\text{l}}_{\text{2}}}{\text{ = molarity }} \times {\text{ L of solution }}

moles of BaCl2 = 0.1M × 0.05L {\text{moles of BaC}}{{\text{l}}_{\text{2}}}{\text{ = 0}}{\text{.1M }} \times {\text{ 0}}{\text{.05L }}

moles of BaCl2 = 0.005 mol{\text{moles of BaC}}{{\text{l}}_{\text{2}}}{\text{ = 0}}{\text{.005 mol}}

Now, using the stoichiometric ratio ofBaCl2{\text{BaC}}{{\text{l}}_{\text{2}}}to Cl - {\text{C}}{{\text{l}}^{\text{ - }}} calculate the moles of chloride ions.

1 mol BaCl2 = 2 mol Cl - 1{\text{ mol BaC}}{{\text{l}}_{\text{2}}}{\text{ = 2 mol C}}{{\text{l}}^{\text{ - }}}

So, 0.005 mol BaCl2 × 2 mol Cl - 1 mol of BaCl2 = 0.01 mol of Cl - {\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:

total moles of Cl - = 0.005 mol + 0.01 = 0.015 mol Cl - {\text{total moles of C}}{{\text{l}}^{\text{ - }}} = {\text{ 0}}{\text{.005 mol + 0}}{\text{.01 = 0}}{\text{.015 mol C}}{{\text{l}}^{\text{ - }}}

total volume of solution = 0.05L + 0.05L = 0.1 L{\text{total volume of solution }} = {\text{ 0}}{\text{.05L + 0}}{\text{.05L = 0}}{\text{.1 L}}

[Cl - ] = 0.015 mol Cl - 0.1 L{\text{[C}}{{\text{l}}^{\text{ - }}}{\text{] = }}\dfrac{{{\text{0}}{\text{.015 mol C}}{{\text{l}}^{\text{ - }}}}}{{0.1{\text{ L}}}}

[Cl - ] = 0.15 M{\text{[C}}{{\text{l}}^{\text{ - }}}{\text{] = 0}}{\text{.15 M}}

So, the molarity of chloride in the mixture of 50 ml50{\text{ ml}} of 0.1 M NaCl0.1{\text{ M NaCl}} and 50 ml50{\text{ ml}} of 0.1 M BaCl20.1{\text{ M BaC}}{{\text{l}}_{\text{2}}}is0.15 M{\text{0}}{\text{.15 M}}.

Thus, the correct option is (C) 0.15 M{\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.