Question

Given the following information: 1.6 g of an unknown monoprotic acid (HA) required 50.80 mL of a 0.35 M NaOH solution to reach the equivalence point calculate the molar mass (g/mol) of the acid. Enter the value ONLY. Do not include the units,

Answer 1

Given: Mass of acid = 1.6 g

Molarity of NaOH = 0.35 M

Volume of NaOH = 50.80 ml or (50.80/1000) L

Molar mass of Acid = ?

At the equivalence point,

No of moles of Acid (HA) = No of moles of Base (NaOH)

As Molarity = No of moles/Volume in litres

So, No of moles = Molarity * Volume (in litres)

=> M * V (acid, HA) = M* V (base, NaOH)

or, No of moles of acid = M (NaOH) * V (NaOH)

As No of moles = Mass/Molar mass

=> Mass of acid/Molar Mass of acid = M (NaOH) * V (NaOH)

=> Molar mass of acid = Mass of acid/ [ M(NaOH) * V (NaOH) ]

Putting the values, we get

Molar mass of acid = 1.6g * 1000 / [ 0.35M* 50.80 L)

Molar mass of acid = 89.99 g/mol

So, Molar mass of Acid = 89.99

for the neutralisation at equivalence pánt, ( No of Moles of Acid = No of Moles of Base As, Molarity = No of Moles Volume of solution lin litres) - No of Moles - Molarity & Volume (M x V) Acid HA = (MXV) Naoh or, No of Moles CHA) = 0.35 MX 50.80 L 1000 Moles = Mass Molar Mass = 0.35 M X 50.80 L Mass of Acid (HA) Molar Mass of Acid 1000 = 1.6 g Molar Mass of Aůd 0.35 M X 50.80 L 1000 Molar Mass of Acid = 1.6 x 1000 0.35 x 50.80 g/mol - Molar Mass of Aid = 89.99 g/mol