Question

A is a solution containing 5.00 g of HNO3 in 500 cm\(^3) of solution. B is a solution of NaOH of unknown concentration. 21.30cm3 of A was titrate it with 25.0 cm3 portions of B using methyl orange as an indicator. calculate the: (i) concentration ot A In mol dm−3
(ii) concentration of B in mol dm−3.

(iii) concentration of B in gdm−3

Answer 1

molar mass HNO3 = approximately 63 g/mol.
molar mass NaOH = approximately 40 g/mol.
mols HNO3 = g/molar mass = 5.00/63 = 0.0794
1. Molarity HNO3 = moles/L = 0.0794/0.500 = 0.159 M = 0.159 moles/dm^3
2. HNO3 + NaOH ==> NaNO3 + H2O
mols HNO3 = M x L = 0.159 M x 0.0213 L = 0.00337
From the equation you see 1 mol HNO3 requires 1 mol NaOH; therefore, 0.00337 mols HNO3 will require 0.00339 mols NaOH. M NaOH = moles/L = 0.00339/0.025 L = 0.135 M = 0.135 mol/dm^3
3. 0.135 mol/dm^3 x 40 g/mol = ? g/dm^3

Answer 2

Yes