Suppose of potassium nitrate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

0.0269 M

Explanation:

There is some info missing. I think this is the original question.

Suppose 0.816 g of potassium nitrate is dissolved in 300 mL of a 14.0 mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

The molecular equation corresponding to this reaction is:

2 KNO₃ (aq) + Na₂CrO₄ (aq) ⇄ K₂CrO₄ (aq) + 2 NaNO₃ (aq)

The full ionic equation is:

2 K⁺ (aq) + 2 NO₃⁻ (aq) + 2 Na⁺ (aq) + CrO₄²⁻ (aq) ⇄ 2 K⁺ (aq) + CrO₄²⁻ (aq) + 2 Na⁺ (aq) + 2 NO₃⁻ (aq)

As we can see, the moles of K⁺ are equal to the initial moles of KNO₃. The molar mass of KNO₃ is 101.10 g/mol. The moles of KNO₃ (and K⁺) are:

0.816 g * (1 mol / 101.10 g) = 8.07 * 10⁻³ mol

The molarity of K⁺ is:

8.07 * 10⁻³ mol / 0.300 L = 0.0269 M