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

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

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Our tutors have indicated that to solve this problem you will need to apply the Weak Base Strong Acid Titrations concept. You can view video lessons to learn Weak Base Strong Acid Titrations. Or if you need more Weak Base Strong Acid Titrations practice, you can also practice Weak Base Strong Acid Titrations practice problems.

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Based on our data, we think this problem is relevant for Professor Eroy-Reveles' class at UCSC.

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.