Question

What is the molarity of calcium bicarbonate if 9.52 mL of 1.20 M HNO3 is required in a titration to neutralize 50.0 mL of a solution of Ca(HCO3)2?

Answer 1

2 HNO3 + Ca (HCO3)2 is Ca(NO3)2+2 CO2+2 H2O
(9.870ml)x(1.000 M HNO3) x (1mol Ca ( HNO3) / (50.00 ml Ca(HCO3)2)= 0.09870 M Ca(HCO3)2

Answer 2

The molarity of calcium bicarbonate if 9.52 mL of 1.20 M HNO3 is required in a titration to neutralize 50.0 mL of a solution of Ca(HCO3)2 is 0.228M.

HOW TO CALCULATE MOLARITY:

• The molarity of a substance in a titration experiment can be calculated by using the following formula:

C1V1 = C2V2

Where;

1. C1 = concentration of acid (M)
2. V1 = initial volume of acid (mL)
3. C2 = concentration of base (M)
4. V2 = volume of the base (mL)

According to this question;

1. V1 = 9.52ml
2. V2 = 50.0ml
3. C1 = 1.20M
4. C2 = ?

1.20 × 9.52 = C2 × 50

11.424 = 50C2

C2 = 11.424 ÷ 50

C2 = 0.228M

• Therefore, the molarity of calcium bicarbonate if 9.52 mL of 1.20 M HNO3 is required in a titration to neutralize 50.0 mL of a solution of Ca(HCO3)2 is 0.228M.

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