Question

3. ) if the calculated molarity of the NaOH solution was lower than the true molarity, would the % acetylsalicylic acid (ASA) in a spring be higher, lower or equal to the actual percent? (explain)

4.) if the molarity of the NaOH solution that you obtained was lower than the true molarity, would the molar mass if the unknown acid that you calculated be higher lower or equal to the actual molar mass? (explain)

Answer 1

3) we have acid base reaction as follows

Acid + base --> salt + water

hence NaOH ( base) if calculated lower molarity it indicates our caluclation of ASA ( acid) will be also lower than true value i.e actual percent of acid

4) since we calculated lower Molarity of acid due to lower NaOH it indicates we calculated lower moles of acid

since Moles = Molarity x Volume of solution , hence lower is molarity the lower is moles we calculate

Now we have formula Moles = mass / Molar mass

we see moles is inversely proportional to Molar mass

hence lower is the moles the higher will be Molar mass calculated for same mass we take.

( example let mass be 1g , if we calculate moles 0.1 then molar mass = 10 g/mol , if moles were less i.e 0.05

then molar mass = 20 g/mol)

Hence overall if NaOH molarity is calculated lower than true value then Molar mass of acid is calculated higher than actual value