Question

What is the concentration in molarity of S2O32- (aq) in a solution prepared by mixing 150 mLmL of 0.149 MM Na2S2O3 (aq) with enough KCl (aq) (HINT: KCl concentration doesn't matter) to make a 250.0 mL solution? Report your answer to 3 significant figures.

Answer 1

Answer:

0.0890 M

Explanation:

Since the concentration of KCl is irrelevant in this case, the concentration of Na2S2O3 can be determined using a simple dilution equation:

C1V1 = C2V2, where C1 = 0.149 M, V1 = 150 mL, V2 = 250 mL

C2 = 0.149 x 150/250

                    = 0.089 M

To determine the concentration of  S2O32- (aq), consider the equation:

$Na_2S_2O_3 => 2Na^+_{(aq)} + S_2O^2^-_3_{(aq)}$

The concentration of Na2S2O3 and S2O32- (aq) is 1:1

Hence, the concentration in molarity of S2O32- (aq) is 0.089 M.

To 3 significant figures = 0.0890 M