6. 0-1 points Gvsuchernt 16v2 14 j3.P001 Weak Base Titration During a titration

Question

6. 0-1 points Gvsuchernt 16v2 14 j3.P001 Weak Base Titration During a titration of a weak base with a strong acid, you are slowing converting molecules of the weak base into ma acid. For the hypothetical weak base, B we see the following B (aq) + H30+ (aq) BH + (aq) + H2O (1) In the problem below you will be adding some strong acid, but not enough to reach the endpoint of the titration. 2.00 mL of hydrochloric acid added to the weak base 1.60M hydrochloric acid in the buret as the titrant ( 18.0 mL of original 1.30 M CSHSN (pyridine) sample You are titrating 18.00 mL of a 1.30 M solution of CsHsN (pyridine) with a strong acid. If you add 2.00 mL of a 1.60 M solution of hydrochloric acid, what is the final pH of the remaining weak base solution? The K, for CsHiyN (pyridine) is 1.7 X 109 final pH -

Explanation / Answer

18 mL of 1.30 M solution of pyridine contains (18 * 1.30)/ 1000 mole = 0.0234 mole pyridine

2 mL of 1.60 M solution of pyridine contains (2 * 1.60)/ 1000 mole = 0.0032 mole HCl

The neutralisation reaction between HCl and pyridine is: HCl + C5H5N = C5H6N+ + -OH

So, 0.0032 mole HCl neutralises 0.0032 mole pyridine and rest (0.0234 - 0.0032) mole = 0.0202 mole of pyridine remains unreacted. Total volume of solution = (18+2) mL = 20 mL.

Concentration of unreacted pyridine is = (0.0202 *1000)/ 20 (M) = 1.01 (M)

[HO- ] = sqrt(Kb* concentration of pyridine) = sqrt (1.7 * 10-9 *1.01) (M) = 4.14 * 10-5 (M)

pOH = -log[OH- ] = -log (4.14 * 10-5) = 4.38

pH = `14-4.38 = 9,62

final pH = 9.62

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