Ap 2005 chemistry scoring guidelines - College Board

[Pages:19]AP? Chemistry 2005 Scoring Guidelines

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AP? CHEMISTRY 2005 SCORING GUIDELINES

Question 1

HC3H5O2(aq) R C3H5O2?(aq) + H+(aq)

Ka = 1.34 ? 10 ? 5

Propanoic acid, HC3H5O2, ionizes in water according to the equation above.

(a) Write the equilibrium-constant expression for the reaction.

Ka =

[H+ ][C3H5O2- ] [HC3H5O2 ]

Notes: Correct expression without Ka earns 1 point. Entering the value of Ka is acceptable.

Charges must be correct to earn 1 point.

One point is earned for the correct equilibrium expression.

(b) Calculate the pH of a 0.265 M solution of propanoic acid.

HC3H5O2(aq) R C3H5O2?(aq) + H+(aq)

I 0.265

0

~0

C

?x

+x

+x

E 0.265 ? x

+x

+x

Ka =

[H+ ][C3H5O2- ] = [HC3H5O2 ]

( x)( x) (0.265 - x)

Assume that 0.265 ? x 0.265, then 1.34 ? 10?5 = x2

0.265 (1.34 ? 10?5)(0.265) = x 2

3.55 ? 10? 6 = x 2 x = [H+] = 1.88 ? 10-3 M pH = ? log [H+] = ? log (1.88 ? 10-3)

pH = 2.725

One point is earned for recognizing that [H+] and [C3H5O2- ] have the same value in the equilibrium

expression.

One point is earned for calculating [H+].

One point is earned for calculating the correct pH.

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AP? CHEMISTRY 2005 SCORING GUIDELINES

Question 1 (continued)

(c) A 0.496 g sample of sodium propanoate, NaC3H5O2 , is added to a 50.0 mL sample of a 0.265 M solution of propanoic acid. Assuming that no change in the volume of the solution occurs, calculate each of the following.

(i) The concentration of the propanoate ion, C3H5O2-(aq) in the solution

mol NaC3H5O2

=

0.496 g

NaC3H5O2 ?

1 mol NaC3H5O2 96.0 g NaC3H5O2

mol NaC3H5O2 = 5.17 ? 10-3 mol NaC3H5O2 = mol C3H5O2-

[C3H5O2-] =

mol C3H5O2- = volume of solution

5.17 ?10-3 mol C3H5O2- 0.050 L

= 0.103 M

One point is earned for calculating the number of

moles of NaC3H5O2.

One point is earned for the molarity of the solution.

(ii) The concentration of the H+(aq) ion in the solution

HC3H5O2(aq) R C3H5O2?(aq) + H+(aq)

I 0.265

0.103

~0

C

?x

+x

+x

E 0.265 ? x 0.103 + x

+x

Ka =

[H+ ][C3H5O2- ] = [HC3H5O2 ]

(x)(0.103 + x) (0.265 - x)

Assume that 0.103 + x 0.103 and 0.265 - x 0.265

Ka = 1.34 ? 10-5 =

( x)(0.103) 0.265

x = [H+] = (1.34 ? 10-5) ? 0.265 = 3.45 ? 10-5 M 0.103

One point is earned for calculating the value of [H+].

The methanoate ion, HCO2-(aq) , reacts with water to form methanoic acid and hydroxide ion, as shown in the following equation.

HCO2-(aq) + H2O(l) R HCO2H(aq) + OH?(aq)

(d) Given that [OH-] is 4.18 ? 10-6 M in a 0.309 M solution of sodium methanoate, calculate each of the following.

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AP? CHEMISTRY 2005 SCORING GUIDELINES

Question 1 (continued)

(i) The value of Kb for the methanoate ion, HCO2-(aq)

HCO2-(aq) + H2O(l) R HCO2H + OH?(aq)

I 0.309

-

0

~0

C ?x

-

+x

+x

E 0.309 ? x

-

+x

+x

x = [OH?] = 4.18 ? 10-6 M

Kb =

[OH- ][HCO2H] [HCO2- ]

=

( x)( x) (0.309 - x)

=

(4.18 ?10-6 )2 (0.309 - x)

x is very small (4.18 ? 10-6 M), therefore 0.309 ? x 0.309

Kb =

(4.18 ?10-6 )2 0.309

= 5.65 ? 10-11

One point is earned for substituting 4.18 ? 10-6 for both [OH-] and [HCO2H], and for calculating the value of Kb.

(ii) The value of Ka for methanoic acid, HCO2H

Kw = Ka ? Kb

Ka =

Kw K b

=

1.00 ?10-14 5.65 ?10-11

Ka = 1.77 ? 10-4

One point is earned for calculating a value of Ka from the value of Kb determined in part (d)(i).

(e) Which acid is stronger, propanoic acid or methanoic acid? Justify your answer.

Ka for propanoic acid is 1.34 ? 10?5, and Ka for methanoic acid is 1.77 ? 10? 4. For acids, the larger the value of Ka , the greater the strength; therefore methanoic acid is the stronger acid because

1.77 ? 10? 4 > 1.34 ? 10?5.

One point is earned for the correct choice and explanation based on the Ka calculated for methanoic

acid in part (d)(ii).

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AP? CHEMISTRY 2005 SCORING GUIDELINES

Question 2 Answer the following questions about a pure compound that contains only carbon, hydrogen, and oxygen. (a) A 0.7549 g sample of the compound burns in O2(g) to produce 1.9061 g of CO2(g) and

0.3370 g of H2O(g).

(i) Calculate the individual masses of C, H, and O in the 0.7549 g sample.

massC

=

1.9061 g CO2 ?

1 mol CO2 44.01 g CO2

?

1 mol C

1

mol

CO2

?

12.01 g C

1 mol C

= 0.5202 g C

massH

=

0.3370 g H2O ?

1 mol 18.016

H2O g H2O

?

2 mol H

1

mol

H2O

?

1.008 g H

1 mol H

= 0.03771 g H

massO = 0.7549 g ? 0.5202 g ? 0.03771 g = 0.1970 g O

Three points total are earned:

One point each for the masses (or moles) of

C, H, and O.

(ii) Determine the empirical formula for the compound.

nC

=

0.5202 g C ?

1 mol C

12.01

g

C

=

0.04331 mol C

nH

=

0.03771 g H ?

1 mol H

1.008

g

H

=

0.03741 mol H

1 mol O

nO

=

0.1970

g

O

?

16.00

g

O

=

0.01231 mol O

0.04331 mol C 0.01231

:

0.03741 mol 0.01231

H

:

0.01231 mol O 0.01231

3.518 mol C : 3.039 mol H : 1.000 mol O

The empirical formula is C7H6O2 .

One point is earned for the number of moles of C, H, and O.*

One point is earned for the empirical formula.*

*based on the three masses as determined in part (a)(i) above

(b) A 0.5246 g sample of the compound was dissolved in 10.0012 g of lauric acid, and it was determined that the freezing point of the lauric acid was lowered by 1.68?C. The value of Kf of lauric acid is 3.90?C m?1. Assume that the compound does not dissociate in lauric acid.

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