Question

An unknown mass of NaOH is dissolved in 500g of H20. This solution is neutralized by adding successively 18g of CH3COOH, 25.2g HNO3, and 29.4g of H2SO4. Determine:

a.) The mass of NaOH

b.)The mass of H20 in the neutral solution

Answer 1

Chemical reaction:

NaOH + CH3COOH --- > CH3COONa + H2O

Moles of CH3COOH = 18g of CH3COOH / 60.05 g/mol

= 0.3 mol CH3COOH

This neutrilize 0.3 mol NaOH and also produce 0.3 mol water.

Chemical reaction:

NaOH + HNO3 --- > NaNO3 + H2O

Moles of HNO3 = 25.2 g of HNO3 / 63.01 g/mol

= 0.4 mol HNO3

This neutrilize 0.4 mol NaOH and also produce 0.4 mol water.

Chemical reaction:

2 NaOH + H2SO4 --- > Na2SO4 +2 H2O

Moles of H2SO4 = 29.4g of H2SO4/ 98.079  g/mol

= 0.3 mol H2SO4

This neutrilize 0.6 mol NaOH and also produce 0.6 mol water.

Now totla mole of NaOH =

0.3+04+0.6 mol NaOH = 1.3 mol NaOH

Amount of NaOH = 1.3 mol NaOH =

= 39.997 g / mol *1.3 mol NaOH

=51.996 g or 52 g NaOH

Now totla mole of H2O =

0.3+04+0.6 mol H2O = 1.3 mol H2O

Amount of H2O = 1.3 mol H2O

= 18.02 g / mol *1.3 mol H2O

=23.426 g or 23.4 g H2O

Total mass of water = 500 g +23.4 g

= 523.4 g H2O