Problem: Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.b. 100.0 mL of 0.10 M C2H5NH2 (Kb = 5.6 x 10 -4) titrated by 0.20 M HNO3

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FREE Expert Solution

At halfway point,

[Base]=[Conjugate Acid]

Using the Henderson Hasselbalch equation:

pH=pKa=-log(10-14Kb)=10.75

100.0 mL C2H5NNH2×0.100 mmol1 mL=10 mmol

At the equivalence point, 10 mmol of HNO3 is added. The strong acid will convert the weak base completely to its conjugate acid:

10 mmol C2H5NH3+×1 mL0.20 mmol=50 mL

Molartiy(M)=10.0 mmol100.0 mL + 50.0 mL=0.067 M

Ka=productsreactants=[C2H5NH2][H3O+]C2H5NH3

Determine the value of Ka:

Ka=KwKb=1.0×10-145.6×10-4=1.786×10-11 

Substituting the equilibrium concentrations:

1.78×10-11=(x)(x)0.067-x

[C2H5NH3]Ka=0.0671.78×10-11>>>500; ignore x in the denominator

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Problem Details

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.

b. 100.0 mL of 0.10 M C2H5NH2 (Kb = 5.6 x 10 -4) titrated by 0.20 M HNO3