Question

To neutralize 100 g of HNO3 solution, 30 cm3 of 15% NaOH solution with a density of 1.15 g / cm3 was used. Calculate the mass fraction (in%) of nitric acid in the solution!

Answer 1

15% of NaOH means

In 100g solution 15g of NaOH

d = m/v

v = m/d = 100/1.15 = 86.95mL

Molarity of NaOH = $\frac{15}{40} × \frac{1000}{86.95}$

=4.31 M

moles of NaOH required = 4.31× $\frac{30}{1000}$

= 0.129moles

moles of HNO₃ needed = 0.129 moles

$\frac{mass}{molar mass}$ = 0.129

mass = 0.129 × molar mass

mass = 0.129×63 = 8.127g

mass fraction = $\frac{8.127}{100} × 100$ = 8.127 %