Question

I need both of these answered assuming for 5 the volume of NAOH is 100 ml and the actual concentration of acetic acid is 20.

(5) What concentration NaOH should be used for your lab if that is what I did with the original vinegar? (choose the best option) (A) 0.0100 M (C) 0.25 (B) 0.100 M (D) 1.00 M (E) 2.00 M (6) Potassium Hydrogen Phthalate (KHP) is used as a primary standard (to standardize basic solutions) because it is solid and air-stable, making it easy to weigh accurately. It's solubility in water is 80 g/L. Convert this to mols dm? Molar mass of KHP is 204.22 g/mol

Answer 1

Original vinegar:

1) Original vinegar typically has 5% acetic acid (ethanoic acid) per litre.

Original vinegar contains 50 gram acetic acid in litre

The molecular weight of acetic acid = 60

Molarity of acetic acid in vinegar = 50g / 60gm^-1

Molarity of acetic acid in vinegar = 0.833 Moles in litre

In order to titrate the 0.833 moles of the acetic acid solution we need an equivalent concentration of NaOH (approximately 1M)

Correct answer: option D

2) Molar mass of KHP = 204.22 g/mol

The solubility of KHP in water in grams = 80g / litre

The solubility of KHP in water in moles = 80/204.22

= 0.39 moles / litre (or) 0.39 mols dm^-3

The solubility of KHP in water in moles =  0.39 mols dm^-3