O ACIDS, BASES AND AQUEOUS EQUILIBRIA Calculating the pH of a weak base titrated
ID: 555834 • Letter: O
Question
O ACIDS, BASES AND AQUEOUS EQUILIBRIA Calculating the pH of a weak base titrated with a strong acid An analytical chemist is titrating 242.2 mL of a 0.6400 M solution of piperidine (C^Hi0NH) with a 0.6000 M solution of HIO3. The p K, of piperidine is 2.89 Calculate the pH of the base solution after the chemist has added 285.4 mL of the HIO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HIO solution added Round your answer to 2 decimal places.Explanation / Answer
we have:
Molarity of HIO3 = 0.6 M
Volume of HIO3 = 285.4 mL
Molarity of C5H10NH = 0.64 M
Volume of C5H10NH = 242.2 mL
mol of HIO3 = Molarity of HIO3 * Volume of HIO3
mol of HIO3 = 0.6 M * 285.4 mL = 171.24 mmol
mol of C5H10NH = Molarity of C5H10NH * Volume of C5H10NH
mol of C5H10NH = 0.64 M * 242.2 mL = 155.008 mmol
We have:
mol of HIO3 = 171.24 mmol
mol of C5H10NH = 155.008 mmol
155.008 mmol of both will react
excess HIO3 remaining = 16.232 mmol
Volume of Solution = 285.4 + 242.2 = 527.6 mL
[H+] = 16.232 mmol/527.6 mL = 0.0308 M
we have below equation to be used:
pH = -log [H+]
= -log (3.077*10^-2)
= 1.5119
PH = 1.51
Answer: PH = 1.51
Feel free to comment below if you have any doubts or if this answer do not work
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