Question

A 11.7 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid.
If 28.6 mL of 6.48×10^-2 M sodium hydroxide are required to neutralize the nitric acid, what is the percent by mass of nitric acid in the mixture?

Answer 1

moles of NaOH reacting = M(NaOH)*V(NaOH)

= 6.48*10^-2 m * 28.6 mL

= 1.853 mmol

= 1.853*10^-3 mol

HNO3 and NaOH react as:

NaOH + HNO3 —> NaNO3 + H2O

so,

moles of HNO3 required = 1.853*10^-3 mol

Molar mass of HNO3,

MM = 1*MM(H) + 1*MM(N) + 3*MM(O)

= 1*1.008 + 1*14.01 + 3*16.0

= 63.018 g/mol

mass of HNO3,

m = number of mol * molar mass

= 1.853*10^-3 mol * 63.018 g/mol

= 0.1168 g

This is mass of HNO3 in sample

% mass = mass of HNO3 * 100 / mass of sample

= 0.1168 *100 / 11.7

= 0.998 %

Answer: 0.998 %