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blagie [28]
3 years ago
12

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

Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

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

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