Question

A sample of 2g of 60% pure chalk (Impurites in the chalk are insoluble in HCl)is dissolved in 250ml of 0.2 (M) HCl.What volume (cm3) of 0.01(N) NaOH is required to neutralize the excess acid?

Answer 1

=> 2 grams of 60 % pure chalk is used.

=> So, 60 % of 2 grams is pure CaCO3 =>60/100 x 2 => 1.2 grams is CaCO3

The first reaction is:

CaCO3 + 2HCl ---------------> CaCl2 + CO2 + H2O

=> Hence, 1 part of Calcium Carbonate reacts with 2 parts of HCl.

Molar mass of CaCO3 = 40 + 12 + 3 x 16 => 40 + 60 => 100 grams per mole.

So, 1.2 grams of CaCO3 is 1.2/100 moles => 0.012 moles of CaCO3.

So, 1 part of CaCO3 = 0.012 moles of CaCO3.

Hence, 2 parts of HCl = 2 x 0.012 => 0.024 moles of HCl.

So, 0.024 moles of HCl react completely.

For HCl solution:

Volume = 250 mL => 0.25 L

Molarity = 0.2 M  

Molarity = (moles of HCl)/(liters of solution)

0.2 = (moles of HCl)/0.25

moles of HCl present = 0.25 x 0.2 => 0.05 moles of HCl

=> So, the solution has 0.05 moles of HCl in it.

Out of this, 0.024 moles of HCl is consumed by CaCO3.

Hence, 0.05 - 0.024 moles of HCl => 0.026 moles of HCl remains.

=>Now, we have to find the volume (in cm3) of 0.01 N NaOH solution required to neutralize 0.026 moles of HCl.

Reaction taking place:

NaOH + HCl ------------> NaCl + H2O

=> Hence, 1 part of NaOH reacts with 1 part of HCl.

So, 1 part of HCl = 0.026 moles of HCl.

Hence, 1 part of NaOH = 1 x 0.026 => 0.026 moles of NaOH.

So, 0.026 moles of NaOH will be required to react with 0.026 moles of HCl.

=> So, the 0.01 N solution should contain 0.026 moles of NaOH.

Normality = 0.01 N

Acidity of NaOH = 1 ( 1 OH-1 ion per molecule of NaOH in aqueous state)

Normality = Molarity x Acidity of Base

0.01 = Molarity x 1

=> Molarity of NaOH solution = 0.01 M.

Now,

Molarity = 0.01 M

Volume of solution = ?

Moles of NaOH = 0.026 moles

Molarity = (moles of NaOH)/(litres of solution)

0.01 = 0.026/(litres of solution)

=> litres of solution = 0.026/0.01 => 2.6 litres of NaOH.

=> Hence, 2.6 litres or 2.6 x 1000 => 2600 mL of 0.01 N NaOH solution will be required.

So the correct answer is 2600 cm3 of 0.01 N NaOH solution.