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

Explanation:

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

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 = $\frac{1}{1}\times 0.015=0.015$ mole of HCl

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

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

$0.130=\frac{0..015}{V_s}$

$V_s=0.115L$

Thus volume required is 0.115 l or 115 ml