This graph shows the titration of 25.0 ml of methylamine solution with 0.0982 m hcl. Determine the molarity of the methylamine solution.

0.0957 M

Explanation:

Equivalence appears to occur upon addition of 20 mls of 0.0957 M HCl.

0.02 L x 0.0957 mole/L = 0.001914 moles HCl added

CH3NH3 + HCl ==> CH3NH4^+ Cl^-

Molarity of CH3NH3 initially present = 0.0957 M

pH < 7 because the reaction is between a strong acid (HCl) and a weak base (methylamine)

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