1. An aqueous solution of nitric acid is standardized by titration with a 0.199

Question

1. An aqueous solution of nitric acid is standardized by titration with a 0.199 M solution of sodium hydroxide. If 15.5 mL of base are required to neutralize 24.6 mL of the acid, what is themolarity of the nitric acid solution? _________M nitric acid

2.An aqueous solution of barium hydroxide is standardized by titration with a 0.177 M solution of perchloric acid. If 18.9 mL of base are required to neutralize 16.0 mL of the acid, what is the molarity of the barium hydroxide solution? _____M barium hydroxide

3.An aqueous solution of hydroiodic acid is standardized by titration with a 0.143 M solution of calcium hydroxide. If 28.8 mL of base are required to neutralize 21.8 mL of the acid, what is the molarity of the hydroiodic acid solution?________ M hydroiodic acid

Explanation / Answer

1. HNO3 + NaOH ------> NaNO3 + H2O

1 mole of HNO3 required 1 mole of NaOH

Therefore,

M1 × V1 = M2 × V2

0.199M × 15.5ml = M2 × 24.6ml

M2 = 0.199M ×15.5ml/24.6ml=0.125M

Therefore , Molarity of HNO3 = 0.125M

2) Ba(OH)2 + 2HClO4 -------> Ba(OCl4)2 + 2H2O

1 mole of Ba(OH)2 require 2 mole of HClO4

N1V1 = N2V2

N1 = Normality of HClO4 = 0.177N

V1 = 16ml

N2 = ?

V2 = 18.9ml

N2 = 0.177M

N2 = 0.177N × 16/18.9

N2 = 0.1498

For Ba(OH2) , Molarity = Normality/2= 0.1498/2= 0.0749M

Therefore, Molarity of Ba(OH)2 = 0.0749M

2) Ca(OH)2 + HI ------> CaI2 + 2H2O

Normality of Ca(OH)2 ,N1= 0.143/2 = 0.0715N

Volume of Ca(OH)2,V1 = 28.8ml

Normality of HI ,N2= ?

Volume of HI ,V2 = 21.8ml

N2 = 0.0715N × 28.8ml/21.8ml = 0.0945N

For HI , Normality = Molarity

Therefore, Molarity of HI = 0.0945M

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