5· The titration curve is shown below 13 (A) a strong acid is titrated with a st

Question

5· The titration curve is shown below 13 (A) a strong acid is titrated with a strong base (B) a weak acid is titrated with a strong base (C) a strong acid is titrated with a weak base (D) a weak base is titrated with a strong acid (E) a strong base is titrated with a weak acid (i) What type of titration is it? 12 10 PH 8 (i) Which one of the following best characterizes the solution in the flask at the equivalence point of titration? (A) a solution of a strong acid (B) a solution of a strong base (C) a solution of a weak acid (D) a solution of a weak base (E) a neutral solution 0 10 Titrant (mL) 15 20 (iii) The equivalence point is at volume of mL and pH of (iv) The pKa of the weak base (or acid) is equal to , based on the curve. 6. What is the molarity of 10.0 mL Ba(OH)2 which can be completely neutralized by 25.0 mL of 0.100 M HCI? The concentration of a solution of a salt of caffeine (a weak base) and hydrochloric acid (HCl) is 0.50 M. The Kb of caffeine is 4.1 x 104. What is the pH of the salt solution?

Explanation / Answer

Q5

i)

This must be a weak acid + strong base

since there is:

- Buffer formatino in the 0-10 mL area

- pH is acidic initially

ii)

the equivalence point will have:

hydrolysis of a conjguate

this is a weak acid which produces conjugate base in equilbirium

so this will be a weak base

choose D

iii)

equivalence point is given at V = 10 mL, since drastic pH change is present htere

pH is approx --> 9 which is the average between max/min V

iv)

pKa can be obtained via

p H= pKa + log(a-/HA)

in the halff (50%) equivalence point, the pH = pK a+ log(1)

so

pH = pKa

approx, V = 10/2 = 5 mL

pH = 5.5

pKA= 5.5

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