Answer: Molarity of chloride anion = 0.32 M
<em>Note: the question is missing some values. The full question is given below;</em>
<em>Suppose 7.26 g of nickel(II) chloride is dissolved in 350 mL of a 0.50 M aqueous solution of potassium carbonate. Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the nickel(II) chloride is dissolved in it. Be sure your answer has the correct number of significant digits.</em>
Explanation:
Molarity or molar concentration is the number of moles (mol) of component per volume (liters) concentration of solution in mol/L or M
The mass of nickel (II) chloride is 7.26 g.
The volume of potassium carbonate is 350 mL = 0.35 L
The molarity of potassium carbonate solution is 0.50 M
The reaction of nickel (II) chloride and potassium carbonate is given below.
NiCl₂(aq) + KCO₃(aq) --------> KCl(aq) +NiCO₃(s)
The dissociation of nickel (II) chloride is given below.
NiCl₂ -----> Ni²⁺ + 2Cl⁻
The molar mass of nickel (II) chloride is 129.6 g/mol
The moles of nickel (II) chloride can be calculated by the formula given below;
No of moles = mass(g) / molar mass (g/mol)
No of moles = 7.26 / 129.6 = 0.056 moles
Therefore, molarity of NiCl₂ = 0.056 moles/ 0.35 L = 0.16 M
The molarity of 1 mole nickel (ii) chloride is 0.16 m and according to dissociation of nickel (II) chloride, 1 mole of nickel (II) chloride gives 2 moles of chloride anion.
Therefore, the molarity of chloride anion = 0.16 * 2 = 0.32 M
