Question

What volume of 0.130 M HCl is required for the complete neutralization of 1.30 g of NaHCO3 (sodium bicarbonate)?

Answer 1

Answer: Volume required is 0.115 L or 115 ml

Step-by-step explanation:

moles of
`NaHCO_3` =
`\frac{\text {given mass}}{\text {Molar mass}}=(1.30g)/(84g/mol)=0.015mol`

The balanced chemical equation is:

`HCl+NaHCO_3\rightarrow NaCl+H_2CO_3`

1 mole of
`NaHCO_3` requires = 1 mole of HCl

Thus 0.015 mol of
`NaHCO_3` requires =
`(1)/(1)* 0.015=0.015` mole of HCl

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

`Molarity=(n)/(V_s)`

where,

n = moles of solute

`V_s` = volume of solution in L

`0.130=(0..015)/(V_s)`

`V_s=0.115L`

Thus volume required is 0.115 l or 115 ml