What is the concentration of sodium chloride in the final solution if 25.34 mL of 0.113 M BaCl2 completely reacts and the total volume of the reaction is 110.4 mL, given the reaction: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)

The concentration of sodium chloride in the final solution is 0.0519 mol/L

Explanation:

In this question, we are asked to calculate the concentration of sodium chloride in the final solution given the volume of Barium chloride and its concentration reacted, including the total volume of the reaction.

Firstly, we proceed with calculating the number of moles of barium Chloride present. This can be calculated by multiplying the molarity of the barium chloride by the volume of the barium chloride.

From the question, we can identify that the volume is 25.34mL and the molarity is 0.113M.

The number of moles is thus; 0.113 * 25.34/1000 = 2.86 * 10^-3 mole BaCl2

Kindly note that we divided the volume by 1000 because we have to convert the milliliter to its standard liter unit.

Now, we need to calculate the number of moles of sodium chloride. From the equation of reaction, we can see that one mole barium chloride yielded 2 moles Sodium chloride.

This means that 2.86*10^-3 mole BaCl2 will yield 2.86 * 10^-3 * 2 NaCl = 5.72 * 10^-3 moles NaCl

Now, we proceed to calculate the concentration of the NaCl since we have the number of moles of it already.

The concentration is simply, the number of moles divided by the final volume.

Mathematically, the concentration of NaCl is;

5.72 * 10^-3/ (110.4/1000) = 0.0519 mol/L

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