Figure 1 of 1 Exercise 18.64 Consider the following curve (Figure 1) for the tit

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Figure 1 of 1

Exercise 18.64

Consider the following curve (Figure 1) for the titration of a weak base with a strong acid and answer each of the following questions.

Part A

What is the pH at the equivalence point?

Express your answer as a whole number.

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Part B

What is the volume of added acid at the equivalence point?

Express your answer using two significant figures.

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Part C

At what volume of added acid is the pH calculated by working an equilibrium problem based on the initial concentration and Kb of the weak base?

Express your answer using one significant figure.

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Part D

At what volume of added acid does pH=14pKb?

Express your answer using two significant figures.

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Part E

At what volume of added acid is the pH calculated by working an equilibrium problem based on the concentration and Ka of the conjugate acid?

Express your answer using two significant figures.

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Exercise 18.64

Consider the following curve (Figure 1) for the titration of a weak base with a strong acid and answer each of the following questions.

Part A

What is the pH at the equivalence point?

Express your answer as a whole number.

pH =

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Part B

What is the volume of added acid at the equivalence point?

Express your answer using two significant figures.

V =   mL  

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Part C

At what volume of added acid is the pH calculated by working an equilibrium problem based on the initial concentration and Kb of the weak base?

Express your answer using one significant figure.

V =   mL  

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Part D

At what volume of added acid does pH=14pKb?

Express your answer using two significant figures.

V =   mL  

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Part E

At what volume of added acid is the pH calculated by working an equilibrium problem based on the concentration and Ka of the conjugate acid?

Express your answer using two significant figures.

V =   mL  

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14 12 10 4 0 10 20 30 40 50 Volume of acid added (mL)

Explanation / Answer

pH in equivalence point...

first, identify the equivalence point

this is the point in which there is drastic pH change

from the graph

this can be seen approx at

pH = 5

b)

The volume of acid required

is given when we find pH = 5

then, V = 25 mL approx

c)

this is when V = 0

since there is only base, and initial concentration is unaffected

d)

pH = 14 - pKb

this is given only in the HALF equivalence point

when there is buffer

that is, when

pOH = pKb

or

pH = 14-pOH = 14-pKb

then,

V half = 1/2*25 = 12.5 mL approx

e)

the only point in which we need to use Ka of conjugate is at equivalence point

it is, when V = 25 mL

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