Question

10. 100.0 mL of a 0.050 mol/L solution of sodium sulfate is added to 150.0 mL of a 0.100 mol/L solution of sodium iodide. What is the new concentration of sodium ions? a. 0.080 M b. 0.10 M c. 0.13 M d. 0.15 M e. 0.25 M

          10. 100.0 mL of a 0.050 mol/L solution of sodium sulfate is added to 150.0 mL of a 0.100 mol/L solution of sodium iodide. What is the new concentration of sodium ions?
a. 0.080 M
b. 0.10 M
c. 0.13 M
d. 0.15 M
e. 0.25 M

Added by Ryan E.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert David Collins

06/08/2023

Step 1

Calculate the moles of sodium ions from sodium sulfate: Moles = Molarity × Volume = 0.050 mol/L × 0.100 L = 0.005 mol Show more…

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10. 100.0 mL of a 0.050 mol/L solution of sodium sulfate is added to 150.0 mL of a 0.100 mol/L solution of sodium iodide. What is the new concentration of sodium ions? a. 0.080 M b. 0.10 M c. 0.13 M d. 0.15 M e. 0.25 M