Question

A 0.07420 gram sample of methyl amine was mixed with 125.0 mL of a HCl solution. The back-titration of the unreacted acid required 5.35 mL of 0.1057 M KOH to reach the equivalence point. Determine the molarity of the acid.

Answer 1

The unreacted acid requires 5.35 mL of 0.1057 M KOH.

So, that means moles of KOH required to neutralize the acid are Molarity x Volume (in L) = 0.1057 M x (5.35 x 10^-3) = 5.65 x 10^-4 mol

HCl + KOH --------> KCl + H2O.

HCl and KOH react in equimolar ratio.

Thus, 5.65 x 10^-4 mol of KOH are required to neutralize 5.65 x 10^-4 mol of HCl.

Thus, moles of unreacted HCl = 5.65 x 10^-4 mol

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Given, mass of methyl amine (CH3NH2) = 0.07420 g

Molar mass of methyl amine = 31.05 g/mol

Moles of methyl amine = Mass / Molar mass = 0.07420 g / 31.05 g/mol = 2.389 x 10^-3 mol

CH3NH2 + HCl ------ CH3NH3^+1 + Cl^-1

Thus, here also they react at equimolar concentration.

Thus, moles of HCl reacted = 2.389 x 10^-3 mol

Total number of moles of HCl = moles of HCl reacted + moles of HCl unreacted

= 5.65 x 10^-4 mol + 2.389 x 10^-3 mol

= 2.9546 x 10^-3 mol

Molarity of acid = Moles of acid / Volume of acid (in L)

= 2.9546 x 10^-3 mol / 0.125 L

= 0.0236 M