Question

This graph shows the titration of 25.0 mL of methylamine solution with a 0.0798 M HCl. Determine the molarity of the methylamine solution.

Answer 1

Clearly from the graph it is evident that equivalence happens at Volume of HCl = 21 mL (Steepest Part of Curve)

So, we have the formula,

    M₁V₁ = M₂V₂

=>M₁ * 25.0 = 0.0798 * 21

=> M₁ = 0.0670 M

So, molarity of methyl amine is 0.0670 M

Answer 2

Answer:

IF:

CH₃NH₂ + HCl ------> CH₃NH₄⁺Cl^-

1moL CH₃NH₂ + 1moL HCl ------> 1moL CH₃NH₄⁺Cl^-

1moL x 31 g/moL CH₃NH₂ + 1moL x 36.45 g/moL HCl ------> 1moL x 67.45 g/moL CH₃NH₄⁺Cl^-

31 g CH₃NH₂ + 36.45 g HCl ------> 67.45 g CH₃NH₄⁺Cl^-

AND IF:

for HCl_sln:

V_HCl= 17 mL = 0.017 L

[M]= 0.0798 M = 0.0798 moL/L

F_w= 36.45 g/moL

Mass_HCl=0.0798 moL/L x 36.45 g/moL x 0.017 L = 0.04945 g pure HCl

for CH₃NH_2sln:

V_CH3NH2= 25 mL = 0.025 L

[M]= X M = X moL/L

F_w= 31 g/moL

THEN:

0.04945 g pure HCl x (31 g pure CH₃NH₂ / 36.45 g pure HCl) = 0.04205 g CH₃NH₂

[M]= Mass_CH3NH2 / (V_sln x F_wCH3NH2)

[M]= 0.04205 g / (0.025 L x 31 g/moL)

[M]= 0.05426 M