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Question: 1mL of 10mM \(KCl\) solution was mixed with 4mL of 0.05 M \(BaC{{l}_{2}}\) solution. The final \(C{{...

1mL of 10mM KClKCl solution was mixed with 4mL of 0.05 M BaCl2BaC{{l}_{2}} solution. The final Cl−C{{l}^{-}} (chloride anion) concentration is:
(A)- 41 mM
(B)- 42 mM
(C)- 67 mM
(D)- 82 mM

Explanation

Solution

From, the moles of the ion present in each solution along with the total volume of the solution. The concentration of ions can be obtained from the formula Molarity formula.

Complete step by step solution:
It is given that two solutions, that is, sodium chloride and barium chloride are mixed together. Such that, 1 ml of 10 mM of KClKCl solution mixed with 4 mL of 0.05 M of BaCl2BaC{{l}_{2}} solution.
So, in order to find the chloride concentration in the final solution, we will find the moles of chloride ion in the total volume of the solution as follows:
In KClKCl solution, given the concentration (10 mM) and volume (1 mL) the moles of KClKCl present will be:
Molarity=Moles of KClVolume of solution in LMolarity=\dfrac{Moles\,of\, KCl }{Volume\,of\,solution\,in\,L}
10 ×10−3M=Moles of KCl1×10−3L10\,\times {{10}^{-3}}M=\dfrac{Moles\,of\, KCl }{1\times {{10}^{-3}}L}
Moles ofKCl=10×10−3M×1×10−3L=10×10−6molesMoles\,of KCl=10\times {{10}^{-3}}M\times 1\times {{10}^{-3}}L=10\times {{10}^{-6}}moles
As we know that one molecule of KClKCl has one potassium and one chloride ion. Then, from the stoichiometric ratio of KClKCl, we have, one mole of KClKCl has one mole of Cl−C{{l}^{-}}. So, 10×10−6 moles10\times {{10}^{-6}}\,moles of KClKCl will have 10×10−6 moles10\times {{10}^{-6}}\,molesof Cl−C{{l}^{-}}ions.
Similarly, in BaCl2BaC{{l}_{2}}, one molecule of barium chloride has one barium and two chloride ions. From the stoichiometric ratio, one mole BaCl2BaC{{l}_{2}} has two moles of chloride ion.
Also, the moles of BaCl2BaC{{l}_{2}} are =Molarity of BaCl2×Volume of solution=Molarity\,of\,BaC{{l}_{2}}\times Volume\,of\,solution
=0.05M×4×10−3L=0.2×10−3moles=0.05M\times 4\times {{10}^{-3}}L=0.2\times {{10}^{-3}}moles
The moles of chloride ion in 0.2×10−3 moles0.2\times {{10}^{-3}}\,moles of BaCl2BaC{{l}_{2}}will be (2×0.2×10−3)=0.4×10−3 moles=400×10−6moles(2\times 0.2\times {{10}^{-3}})=0.4\times {{10}^{-3}}\,moles=400\times {{10}^{-6}}moles
Therefore, the moles of chloride ion in the final solution will be obtained by adding the moles in KClKCl and BaCl2BaC{{l}_{2}} solution. So, we get (10×10−6+400×10−6)=410×10−6 moles(10\times {{10}^{-6}}+400\times {{10}^{-6}})=410\times {{10}^{-6}}\,moles.
Also, the volume of the total solution will be (1+4)=5×10−3L(1+4)=5\times {{10}^{-3}}L
Thus, the molarity of the solution will be =MolesVolume=410×10−65×10−3=82×10−3M=82mM=\dfrac{Moles}{Volume}=\dfrac{410\times {{10}^{-6}}}{5\times {{10}^{-3}}}=82\times {{10}^{-3}}M=82mM

Therefore, the final Cl−C{{l}^{-}} (chloride anion) concentration in the solution is option (D)- 82 mM.

Note: The unit conversion must be taken care of, as the concentration (in molarity) is expressed in SI units of moles and litres. Also, the stoichiometric ratio of the molecule should also be considered while calculating the moles of ions in the solution.