Question

A 0. 462 g sample of a monoprotic acid is dissolved in water and titrated with 0. 180 m koh. what is the molar mass of the acid if 28. 5 ml of the koh solution is required to neutralize the sample?

Answer 1

The molar mass of the acid if 28. 5 ml of the koh solution is required to neutralize the sample is 90.23g/mol.

For the calculation of molarity of solution

Molarity = (moles of solute/volume of solution) × 1000

Given,

Molarity of KOH solution = 0.180 M

Volume of solution = 28.5 mL

0.180 = (moles of KOH/ 28.5) × 1000

Moles = (0.18× 28.5)/1000

= 0.00513 mol

Chemical equation for the reaction

HA + KOH ------- KA + H2O

1 moles of KOH reacts with 1 moles of HA.

So, 0.00513 moles of KOH react with 0.00513 moles of HA.

To calculate the molar mass for given number of moles

Number of moles= given mass/ Molar mass

Given,

Mass of HA = 0.462 g

Moles of HA = 0.00512 mol

0.00512 = 0.462/ Molar mass

Molar mass = 90.23 g/ mol.

Thus the molar mass of HA required to neutralize the 28.5 mL of KOH is 90.23g/mol.

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