Answer to Question #345704 in General Chemistry for Kaushi

Question

A 500cm3 of solution is prepared by dissolving 0.25g of CaCo3 in excess HCl.Ca2+ ions concentration in the solution in ppm?

Answer 1

CaCO₃ + 2 HCl = CaCl₂ + CO₂ + H₂O

n (CaCO₃) =0,25g / 100 g/mol = 0,0025 mol;

V = 500sm³ = 500 ml = 0,5 litr.

1mol CaCO3 => 1mol Ca²⁺

0,0025mol => X

X = 0,0025 mol Ca²⁺

C_M = n / V = 0,0025 mol / 0,5 l = 0,005 mol/l

ANSWER: 0,005 mol/l Ca²⁺