Suppose a student made a different sodium hydroxide solution using 0.401g of solid sodium hydroxide and 200mL of water. The student then standardized this solution using benzoic acid in a similar manner to that depicted in the simulation. From the following additional data recorded by the student: a) Mass of benzoic acid: 0.158g b) Volumer of the benzoic acid solution: 100.0mL c) Volume of sodium hydroxide need to neutralize the solution: 27.84mL Determine the molarity of the sodium hydroxide solution.

0.05 M

Explanation:

Mass of benzoic acid= 0.158g

Volume of benzoic acid= 100 mL

Volume of sodium hydroxide = 27.84mL

Molar mass of benzoic acid= 122g/mol

Number of moles of benzoic acid= 0.158g/122g/mol= 1.3 × 10^-3 moles

C= no of moles/volume

C= 1.3 × 10^-3 moles × 1000/100

C= 0.013M

So;

Volume of acid VA = 100mL

Concentration of acid CA= 0.013M

Volume of Base VB = 27.84mL

Concentration of Base CB= ???

Number of moles of acid NA =1

Number of moles of Base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB= CBVBNA

CB= CAVANB/VBNA

CB= 0.013 × 100 × 1/27.84 × 1

CB= 0.05 M

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c, liquids evaporate, creating a gas