*Worksheet says:\" you will need to use the tables of Ka and Kb values to answer

Question

*Worksheet says:" you will need to use the tables of Ka and Kb values to answer some of these questions."

*Table not given on worksheet

40mL of a 0.100MHF solution (in flask) is titrated with a 0.200 M NaOH solution (in the buret). How many mL of the NaOH solution needs to be added to reach the equivalence point of the titration? a) 10.0 mL b) 15.0 mL. c) 30.0 mL d) 20.0 mL. e)60.0 mL Which one of the following best characterizes the solution in the flask after 10.00 mL of the NaOH solution has been added? Circle the correct answer. a) a solution of a strong acid b) a solution of a weak acid c) a buffer solution d) a solution of a strong base e) a solution of weak base f) a neutral solution What will be the pH of the solution in the flask after 10.00 mL of the NaOH solution has been added? a) 3.16 b) 12.45 c)2.20 d) 5.82 e) 3.46 Which one of the following best characterizes the solution in the flask at the equivalence point of the titration? Circle the correct answer, a) a solution of a strong acid b) a solution of a weak acid c) a buffer solution d) a solution of a strong base e) a solution of a weak base f) a neutral solution What is the pH of the solution at the equivalence point of the titration? a) 12.55 b)7.00 c) 6.27 d) 8.14 e) 9.11

Explanation / Answer

B)

Q4.

for equivalence point

mmol of acid = mmol of base

mmol = M*V

so

Macid*Vacid = Mbase*Vbase

0.1*40 = 0.2*Vbase

Vbase = 0.1*40/0.2 = 20 mL

Q5.

since the volume is LESS than that of the equivalence point, then:

we will have a buffer solution, since NaOH converts ssome weak acid into conjugate base, so it is a buffer

Q6

the pH , apply henderson haselbach equations

pH = pKa + log(F-/HF)

pKa = 3.41

so

mmol of NAOH = MV = 10*0.2 = 20

mmol of HF = MV = 400.1 = 40

after reaction

mmol of F- = 0 + 20

mmol o fHF = 40-20 = 20

pH = 3.14+ log(20/20) = 3.14

pH = 3.14

Q7

in equivalence point, there will be

HF + NaOH = NaF + H2O

and NAF --> Na+ + F-

so

F- hydrolysis:

F- + H2O <-> HF + OH-

then, it is slightly basic solution

since F- is a weak base

Q8

pH in equivalence point:

choose betwee 8.14 and 9.11

so:

Kb = [HF][OH-]/[F-]

Kb = (10^-14 ) / (10^-3.14) = 1.3803*10^-11

find concentration of F-:

F- = M1*V1 / Vtotal = (40*0.1)/(40+20) = 0.0666

1.3803*10^-11 = x*x/(0.0666-x)

[OH-] = x= 9.54*10^-7

pOH = -log( 9.54*10^-7) = 6.02

pH = 14-6 = 8 approx

choose D

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