Problem: Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.210 M pyridine, C5H5N(aq) with 0.210 M HBr(aq) (Kb = 1.7 x 10-9)after addition of 32.0 mL of HBr

FREE Expert Solution

HBr (strong acid) will react with C5H5N (weak base).

C5H5N(aq) + HBr(aq) → C5H5NH+(aq) + Br-(aq)


Step 1: Calculate the initial amount (in moles) of each species:

molarity (M)=molL

moles C5H5N=0.210mol C5H5NL 25.00 mL x 10-3 L1 mL

moles C5H5N = 0.00525 mol C5H5N


moles HBr=0.210mol HBrL 32.00 mL x 10-3 L1 mL

moles HBr = 0.00672 mol HBr


Step 2: Construct an ICF chart to determine the final amount of each species.


After the reaction is complete the solution contains:

0.00147 mol HBr → strong acid
0.00525 mol C5H5NH+  conjugate acid
 pH will depend on the strong acid


Step 3: Calculate the pH of the solution 

molarity (M)=molL


final volume of solution=volume C5H5N+volumeHBrfinal volume of solution=25.0 mL x 10-3 L1 mL+32.0 mL x 10-3 L1 mLfinal volume of solution=0.025 L+0.032 L

final volume of solution = 0.057 L

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Problem Details

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.210 M pyridine, C5H5N(aq) with 0.210 M HBr(aq) (Kb = 1.7 x 10-9)

after addition of 32.0 mL of HBr

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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.