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