Question

Two 20.0 mL samples, one 0.200 mol L KOH and the other 6.200 mol L-CH3NH, were titrated with 0.100 mol L-HI. a. What is the volume of added acid at the equivalence point for each titration? b. Predict whether the pH at the equivalence point for each titration will be acidic, basic, or neutral. c. Predict which titration curve will have the lower initial pH. . Make a rough sketch of each titration curve.

Answer 1

a) Volume of sample 20.0 mL, Concentration of KOH = 0.200 mol L-1, Concentration of CH3NH2 = 0.200 mol L-1, Concentration of HI = 0.100 mol L-1,

Volume of acid added for KOH can be calculated by

MKOHVKOH = MHIVHI

(0.200) (20.0) = (0.100) (VHI)

VHI = 4/0.100 = 40 mL

Volume of acid added for CH3NH2 can be calculated by

MCH3NH2 VCH3NH2 = MHIVHI

(0.200) (20.0) = (0.100) (VHI)

VHI = 4/0.100 = 40 mL

b. At the equivalence point, for KOH titration pH will be neutral as the salt is formed after the reaction of strong acid (HI) and strong base (KOH).

At the equivalence point, for CH3NH2 titration pH will be acidic as the salt is formed after the reaction of strong acid (HI) and weak base (CH3NH2).

c. KOH --------> K+ +OH- (HIgh pH and low pOH)

CH3NH2 + H2O <--------> CH3NH3+ + OH - (Low pH and high pOH)

At the starting point of titration, strong base has higher pH and weak base has lower pH. Therefore, weak base i.e., CH3NH2 will have lower initial pH.

d) Refer attached image

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