Question

What is the molarity of calcium bicarbonate if 9.56 mL of 1.30 M HNO3 is required in a titration to neutralize 50.0 mL of a solution of Ca(HCO3)2?

Answer 1

=> 0.12428 M

Step-by-step explanation:

To begin, write down a balanced equation;

Ca(HCO3)2 + 2HNO3 => Ca(NO3)2 + 2H2O + 2CO2

Then calculate the number of moles contained in 9.56 mL of 1.30M HNO3

We know that molarity is contained in 1000mL

1000 mL = 1.30 moles

9.56 mL = ?

= (9.56 × 1.30)/1000

= 0.012428 moles

Mole ratio

Ca(HCO3)2 : HNO3

1 : 2

But, xmoles : 0.012428 moles

x = 1/2 × 0.012428

= 0.006214 moles

This moles is contained in 50 mL solution.

50 mL = 0.006214 moles

1000 mL = ?

= (1000 × 0.006214)/50

= 0.12428M