Question

A solution is prepared by dissolving 5g HNO3 in pure water to a total volume of 1L.

Answer 1

Answer:

a)1.1  b) 3.17 g c) 0.793 L

Explanation:

Balance equation for the reaction is

HNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + H₂O (l)

to calculate the pH;

we need to find the molarity of the acid HNO₃

Molarity = no moles / volume in liters

no mole = mass given / molar mass where molar mass of  HNO₃ = 63.01 g/mol

no moles = 5 / 63.01 = 0.0794 moles

molarity = 0.0794 / 1L = 0.0794 M

a) pH = - log (H⁺)  where hydrogen ion concentration is 0.0794 since 1 mole HNO₃ ionizes to 1 mole of H⁺ concentration,  0.0794 will also ionize to give 0.0794 moles

pH =  log (0.0794)⁻¹ = 1.1

b) using the equation of the reaction

63.01 g of  HNO₃ require 39.991 g of NaOH

1g will need $\frac{39.991}{63.01}$

5g will need  $\frac{39.991}{63.01}$ × 5 = 3.17 g

c) volume of 0.1M of NaOH needed = no of moles of NaOH / Molarity

no of moles of NaOH needed = 3.17 / 39.991 = 0.0793

volume need = 0.0793 / 0.1 = 0.793 L