Question

Suppose 3.81g of nickel(II) chloride is dissolved in 250.mL of a 69.0m 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. Round your answer to 3 significant digits.

Answer 1

Ans :

NiCl2 (s) + K2CO3 (aq) = NiCO3 (s) + 2KCl (aq)

Mol NiCl2 = mass / molar mass

= 3.81 g / 129.5994 g/mol

= 0.029 mol

Mol K2CO3 = molarity x volume (L)

= 0.069 M x 0.250 L

= 0.01725 mol

0.01725 mol K2CO3 will form : 2 x 0.01725 mol = 0.0345 mol KCl

so mol chloride anion in solution = 0.0345 mol

so molarity of chloride anion = mol chloride / volume (L)

= 0.0345 mol / 0.250 L

= 0.138 M