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.

The molar mass of the monoprotic acid would be 89.99 g/mol

First, let us look at the equation of the reaction:

HA + NaOH ---> NaA + H2O

Mole ratio of acid to base = 1:1

Volume of NaOH required = 50.80 mL

Molarity of NaOH = 0.35 M

Mole of NaOH = molarity x volume

                        = 0.35 x 50.80/1000

                             = 0.01778 mole

Since the mole ratio according to the balance equation is 1:1, it means that the mole of the monoprotic acid would also be 0.01778.

Thus, molar mass of the monoprotic acid = mass/mole

                      = 1.6/0.01778

                         = 89.99 g/mol

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89.9

Explanation:-

The balanced chemical equation is :

To calculate the moles for given molarity, we use the equation:

    .....(1)

Molarity of solution = 0.35 M

Volume of solution = 50.80 ml

Putting values in equation 1, we get:

According to stoichiometry:

1 mole of NaOH reacts with = 1 mole of HCl

Thus 0.0178 moles of NaOH reacts with = moles of HCl

mass of HCl =

1.6g=  

Molar mass of HCl = 89.9

The molar mass (g/mol) of the acid is 89.9