Review 17ANS

REVIEW QUESTIONS
Chapter 17

A buffer is prepared by adding 20.0 g of acetic acid and 20.0 g of sodium acetate in enough water to prepare 2.00 L of solution. Calculate the pH of this buffer?

What is the ratio of to in blood of pH 7.4 ? ( for )

How many grams of NaBrO should be added to 1.00 L of 0.200 M HBrO to form a buffer with a pH of 8.80 ? ( for )

Acetylsalicylic acid (aspirin, ) is a weak acid with at of sodium acetylsalicylate is added to 200.0 mL of 0.100 M solution of this acid. Calculate the pH of the resulting solution at .

The equations and dissociation constants for three different acids are given below:

Identify the conjugate pair that is best for preparing a buffer with a pH of 7.2. Clearly explain your choice.

The best conjugate pair would be and The for this buffer when

A sample of 25.0 mL of 0.100 M solution of HBr is titrated with 0.200 M NaOH . Calculate the pH of solution after 10.0 mL of the base is added.


Initial	2.50 mmol	2.00 mmol	0	----
	-2.00 mmol	-2.00 mmol	+2.00 mmol	----
Final	0.50 mmol	0	2.00 mmol	----

A buffer solution is prepared by adding 0.10 L of 2.0 M acetic acid solution to 0.10 L of 1.0 M NaOH solution.
a) Calculate the pH of this buffer solution.


Initial				----
				---
Final				----

b) 0.10 L of 0.20 M HCl is added to 0.40 L of the buffer solution above. What is the pH of the resulting solution?

The

ions provided by HCl react with the acetate ions in the buffer.


Initial	0.20	0.020	0.20	----
	-0.020	-0.020	+0.020	----
Final	0.18	0	0.22	----

A 10.0 mL solution of is titrated with a 0.100 M HCl solution. Calculate the pH of this solution at equivalence point.

At equivalence point

of HCl
At equivalence point all

reacts with all

to produce 1.00 mmol of

. Since only salt is present at this point, the

of solution is based on hydrolysis of this salt.


Initial		----
		----
Equil.		----

A solution of is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the HCl solution: (a) 0.0 mL , (b) 10.0 mL , (c) 30.0 mL a) Since no acid has been added, the of solution is based on the ionization of .

b) Addition of

of acid neutralizes some of the ammonia, as shown below:


Initial				----
				----
Final				----

c) After addition of 30.0 mL of

equivalence point is reached. At this point all

reacts with all

to produce 3.00 mmol of

. Since only salt is present at this point, the

of solution is based on hydrolysis of this salt.


Initial		----
		----
Equil.		----

A sample of 0.200 M acetic acid is titrated with 0.180 M NaOH . Calculate the pH of the solution (a) before addition of NaOH, (b) after addition of 20.0 mL of NaOH and (c) at the equivalence point.
a) Since no base has been added, the of solution is based on the ionization of acid.

b) Addition of

neutralizes some of the acetic acid, as shown below:


Initial	9.00 mmol	3.60 mmol	0	----
	-3.60 mmol	-3.60 mmol	+3.60 mmol	----
Final	5.40 mmol	0	3.60 mmol	----

c) At equivalence point:

At this point all the acid

is neutralized by the base

to produce 9.00 mmol of salt. Since only salt is present, the

of the solution is based on hydrolysis of this salt.


Initial		----
		----
Equil.		----

Calculate the molar solubility of in 0.50 M NaBr solution.


Initial	----
	-x	+x
Final	-----	x

A solution is made by mixing 10.0 mL of and 10.0 mL of 0.0010 M . Will a precipitate form? (Ksp for )

The solubility of iron (II) hydroxide, , is .
a) Calculate the Ksp for iron (II) hydroxide.

b) Calculate pH of a saturated solution of iron (II) hydroxide.

c) A 50.0 mL sample of

solution is added to 50.0 mL of

NaOH solution. Does a precipitate form?

Since

, precipitation will not occur
14. Lead iodate,

, is a slightly soluble salt with a Ksp of

. To 35.0 mL of

solution is added 15.0 mL of

. A precipitate of

results. What are the

and

in the final solution?

Using bounce-back method, first assume all

reacts with all

ion to produce

, and then some of the precipitate dissolves back to the ions.


Initial	----

Precipitate	-----

Finish	-----

Consider a solution that is 0.010 M in and 0.020 M in . If sodium sulfate is added to this solution to selectively precipitate one of the cations, which will precipitate first? What is the minimum concentration of that would trigger the precipitation of this cation? From textbook,

Since the solution stoichiometry for both these compounds are similar, it would be appropriate to relate

and molar solubility. Since the lower

value would require the lower sulfate ion concentration in order to precipitate, it would then follow that

would precipitate first.

What is the concentration when 0.010 mol of is dissolved in a liter of solution buffered at pH of 10.0. forms a complex ion with hydroxide shown below:

Due to the large

all of

is converted to the complex ion, and some subsequently dissociates back to

. Then at equilibrium,

Note: Since the solution is buffered, [

] will remain constant during the reaction.
17. A

sample of

is dissolved in 1.00 L of

. If 0.010 mol of NaCl is added to this solution, will

precipitate?

and

form the complex ion

with

)

To determine if a precipitate forms, we need to determine the concentration of

in the solution, and then calculate

to determine if a precipitate forms.

To determine the concentration of

, due to the large

value, assume all of the

reacts to form the complex and then some dissociates back.


Initial

Complex
	+2 x	-x
Finish

AgCl will NOT precipitate
18.

is added to a solution that is 0.10 M in NaCl and

. Assume no dilution caused by the addition of

. Given the Ksp values below:

a) Which precipitates first, AgCl or

? Calculate the

when precipitation first begins.
Since the solution stoichiometry for both these compounds are not the same, it would not be appropriate to relate

and molar solubility.

To determine which ion precipitates first, we must calculate the cation concentration required for each precipitation. The lower value required for precipitation would indicate the ion that would precipitate first.
[

] required for AgCl precipitation:

required for

precipitation:

The precipitation occurs for the salt which requires the smallest

at equilibrium. Therefore

precipitates first at a

b) What is the

when

first begins to precipitate?

Calculations above show that in order of

to precipitate, the

must equal

. Therefore,

Blood is buffered by system. Normal blood plasma is and . for at body temperature is 6.1.
a) What is pH of blood plasma?

b) If the volume of blood in a normal adult is 5.0 L , what mass of HCl can be neutralized by the buffering system in blood before the pH falls below 7.0 (which would result in death)?

For

to drop by

to below

, the ratio of base to acid in the buffer must change to 8 (from the current 20). The amount of the components of the buffer in 5.0 L of blood are:

The HCl neutralized by the buffer reacts with

to form

. Therefore, after

is neutralized, the amounts of buffer components will be:

Solving for

neutralized):

c) For the same adult in (b), what mass of NaOH can be neutralized before the pH rises above 7.8?

For

to rise by 0.4 to above 7.8 , the ratio of base to acid in the buffer must change to 50 (from the current 20). The NaOH neutralized by the buffer reacts with

to form

. Therefore, after

is neutralized, the amounts of buffer components will be:

Solving for

neutralized):

An important buffer used in biochemical analysis is made by dissolving TRIS in dilute HCl . A biochemist prepares a buffer by dissolving an unknown amount of TRIS in 1L of 0.095 M HCl solution. The pH of the resulting buffer solution was measured to be 8.53 . How many grams of TRIS was used in this buffer? Assume volume of solution did not change after addition of TRIS.
(Molar mass of TRIS = of TRIS = 5.91)

In order to prepare the buffer, all the HCl in solution must react with added TRIS to form the conjugate acid and leave excess TRIS as the weak base. Therefore,

	TRIS	HCl	TRISH Cl
In	x	0.095	0.00
	-0.095	-0.095	+0.095
End	x-0.095	0.00	0.095

Solve for

(mol TRIS added)

REVIEW QUESTIONS Chapter 17

REVIEW QUESTIONS
Chapter 17