A. Consider the following graphs to answer questions 1-3: Which one of the graph

Question

A. Consider the following graphs to answer questions 1-3: Which one of the graphs represents to the titration of a weak acid (in flask) with a strong base (in buret)? ________ Analyze graph (b). What was the original concentration of the acid or base in the flask if the original volume was 100 mL and it was titrated with a standard solution (in buret) that had a concentration of 0.100 M? a.0.100 b 0.150 M c. 0.200 M d. 0.250 M e. 0.300 M Analyze the graph that you consider to represent the titration curve for the neutralization of a weak base with a strong acid. From the data shown, what is the approximate K_b. of the weak base? 1 times 10^-4 b. 1 times 10^-5 c. 1 times 10^-6 d. 1 times 10^-7 e. 1 times 10^-8

Explanation / Answer

Q1.

weak acid initially +storng base addition:

this must be the titration curve with initial pH = acidic, since it has wek acid only, so ignore b and d

for the equivalent point, expect pH = 7 for storng acid + strong base, which is clearly titration c, so

the weak acid is that in titration curve (A)

Q2.

from b)

original concentration of acid/base in flask, if V= 100 mL

and the concentration had = 0.1 M

so

mmol of acid = mmol of base

Macid*Vacid = Mbase*Vbase

volume of acid required = 250 mL approx (for equivalence point)

0.1*250 = 100 * Mbase

Mbase = 0.1*250/100 = 0.25 M, choose D

Q3.

weak base + strong acid...

previously, we stated that only b and d are bases (initially)

so ignore a and c

now... there is a very storng pH change in titraiton b, meaning that this is a strong acid/strong base titration

so the best option of weak base is D

for pKb, we need:

pOH = pKb + log(BH+/B)

in the half equivalence point, i.e. V = 200 mL is equivalence point ---> 1/2V = 100 mL is half equivalence

the pH = 8

meaning that

pOH = 14-pH = 14-8 = 6

so

pOH = pKb = 6

that is

Kb = 10^-pKb = 10^-6

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