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What is the molarity of calcium bicarbonate if 9.15 mL of 1.30 M HNO3 is required in a titration to neutralize 50.0 mL of a solution of Ca(HCO3)2?

User Kiennt
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Answer: In a titration, the reaction between the two substances allows us to determine the unknown concentration of one of the solutions. Let's first write the balanced chemical equation for the reaction between calcium bicarbonate (Ca(HCO3)2) and nitric acid (HNO3):

Ca(HCO3)2 + 2HNO3 → Ca(NO3)2 + 2H2O + 2CO2

From the balanced equation, we can see that 1 mole of calcium bicarbonate (Ca(HCO3)2) reacts with 2 moles of nitric acid (HNO3).

Given the volume and concentration of HNO3 used in the titration:

Volume of HNO3 (V1) = 9.15 mL = 9.15 x 10^(-3) L

Molarity of HNO3 (M1) = 1.30 M

And the volume of calcium bicarbonate solution (V2) used in the titration:

Volume of Ca(HCO3)2 (V2) = 50.0 mL = 50.0 x 10^(-3) L

We can use the stoichiometry of the reaction to find the molarity of calcium bicarbonate (M2):

M1 * V1 = M2 * V2

Substitute the values:

1.30 M * (9.15 x 10^(-3) L) = M2 * (50.0 x 10^(-3) L)

Now solve for M2:

M2 = (1.30 M * 9.15 x 10^(-3) L) / (50.0 x 10^(-3) L)

M2 ≈ 0.237 M

The molarity of the calcium bicarbonate solution (Ca(HCO3)2) is approximately 0.237 M.

User Luca Colonnello
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