You will need to use the Tables of and values to answer some of these questions.

Question

You will need to use the Tables of and values to answer some of these questions. A. Consider the following graphs to answer questions 1-3: Titration (a) Titration (b) Titration (d) Titration (c) 100 200 300 400 1. Which one of the graphs represents to the titration of a weak acid (in flask) with a strong base (in buret)? 2. 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? b 0.150 M d. 0.250 M e. 0.300 M a. 0.100 M c. 0.200 M Show calculations: 3. 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. of the weak base? b. 1x10 c. 1x10 e. 1x10 (a) 1x10 d. 1x10 Show calculations:

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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