Answer to Question #115443 in General Chemistry for Andrw Aitken

Question

200.0 mL of a 0.25 M solution of Sodium sulphate is reacted with a 150.0 mL solution of 0.35 M Barium chloride.

Determine the concentration of all the ions present in the final solution

Answer 1

0.25 x 0.2 = 0.05 moles of Na2SO4.

0.15 x 0.35 = 0.0525 moles of BaCl2.

BaCl2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2 NaCl (aq)

0.1 /0.35 = 0,285714286 M of Nacl.

0.0025/0.35 = 0,00714285714 M. BaCl2.