CHM 2046 Answer Key – Practice Quiz 2

CHM 2046

Answer Key ¨C Practice Quiz 2

Answer all questions. Be careful to give your final answer to the correct sig. figs.

1.

a)

(3 points) The acetate ion (CH3COO-) is a weak base. Write down the

base-ionization equation for CH3COO- in water.

CH3COO- (aq) + H2O (l) ? CH3COOH (aq) + OH- (aq)

b)

(2 points) Complete the expression for the base-ionization constant (Kb)

for CH3COO-.

Kb =

c)

[CH3 COOH][OH - ]

[CH3 COO - ]

(5 points) Calculate the pH of a 0.40 M solution of CH3COO- (Ka for

CH3COOH = 1.8 x 10-5).

I

C

E

Kb

CH3COO- (aq) + H2O (l) ? CH3COOH (aq) + OH- (aq)

0.40

0

0

-x

+x

+x

(0.40-x)

x

x

-14

for CH3COO =Kw/Ka (CH3COOH)=(1.0 x 10 )(1.8 x 10-5) = 5.6 x 10-10

Kb = x2/(0.40-x) ¡Ö x2/0.40 ¡à x = [OH-] = 1.50 x 10-5

¡à pOH = 4.82

¡à pH = 9.18

2.

a)

(3 points) Calculate the pH when 100 mL of 0.100 M Ca(OH)2 solution is

added to 50 mL of 0.400 M HCl solution.

Ca(OH)2 (s) + H2O (l) ¡ú Ca2+ (aq) + 2 OH- (aq)

HCl (g) + H2O (l) ¡ú H3O+ (aq) + Cl- (aq)

n(OH ) = MV = (2)(0.100 M)(100 x 10-3 L) = 0.200 moles

n(H3O+) = MV = (0.400 M)(50 x 10-3 L) = 0.20 moles

¡à OH- and H3O+ neutralize exactly

¡à

pH = 7.00

b)

(2 points)

Which one of the following salts will give a basic solution

when dissolved in water? Circle your choice.

CaCl2, NH4Cl, NaClO4, Na2CO3, KI, none of these

(2 points)

Write an equation for the reaction that occurs when the salt

dissolves in water and makes the solution basic, or state why none do.

CO32- (aq) + H2O (l) ? HCO3- (aq) + OH- (aq)

c)

(2 points)

Which one of the following salts will give an acidic solution

when dissolved in water? Circle your choice.

Ca3(PO4)2, NaBr, FeCl3, NaF, KNO2, none of these

(2 points)

Write an equation for the reaction that occurs when the salt

dissolves in water and makes the solution acidic, or state why none do.

FeCl3 (s) + 6 H2O (l) ¡ú [Fe(H2O)6]3+ (aq)

Then, [Fe(H2O)6]3+ + H2O (l) ? [Fe(H2O)5(OH)]2+ + H3O+ (aq)

3.

(a) (8 points) What is the pH of a buffer solution prepared from

adding 60.0 mL of 0.36 M ammonium chloride (NH4Cl) solution to 50.0 mL of

0.54 M ammonia (NH3) solution? (Kb for NH3 is 1.8 x 10-5).

new volume (Vf) = (60.0 + 50.0) mL = 110.0 mL

MV

(0.36 M)(60.0 mL)

new [NH4+] = i i =

= 0.196 M

110.0 mL

V

f

MV

(0.54 M)(50.0 mL)

new [NH3] = i i =

= 0.245 M

110.0 mL

V

f

+

Ka (NH4 ) = Kw/Kb(NH3) = (1.0 x 10-14)/(1.8 x 10-5) = 5.6 x 10-10

pH = pKa + log

OR

[base]

[acid]

= 9.25 + log

[0.245]

= 9.25 + 0.097 =

[0.196]

9.35

n(NH4+) = MV = (60.0 mL)(10-3L/mL)(0.36 M) = 2.16 x 10-2 mol

n(NH3) = (50.0 mL)(10-3L/mL)(0.54 M) = 2.70 x 10-2 mol

Ka (NH4+) = Kw/Kb(NH3) = (1.0 x 10-14)/(1.8 x 10-5) = 5.6 x 10-10

pH = pKa + log

n(base)

2.70x10 ?2

= 9.25 + log

=

2.16x10 ?2

n( acid)

9.35

OR after calculating new [ ]'s in the first part:

NH3 (aq) + H2O(l) ? NH4+ (aq) + OH- (aq)

0.245 M

0.196 M

0

-x

+x

+x

(0.245-x)

(0.196+x)

x

0.196x

x(0.196 + x)

¡Ö

¡à x = 2.25 x 10-5 M = [OH-]

Kb = 1.8 x 10-5 =

0.245

(0.245 ? x)

¡à pOH = -log 2.25 x 10-5 = 4.65

¡à pH = 9.35

(b) (13 points) What is the final pH after 10.0 mL of 0.200 M NaOH

solution is added to a 50.0 mL solution of 0.400 M acetic acid (CH3COOH)?

Ka for acetic acid is 1.8 x 10-5.

new volume (Vf) = (10.0 + 50.0) mL = 60.0 mL

n(CH3COOH) = MV = (0.400 M)(50.0 mL(10-3 L/mL) = 20.0 x 10-3 mol

n(OH- added) = MV = (0.200 M)(10.0 mL)(10-3 L/mL) = 2.00 x 10-3 mol

On mixing, 2.00 x 10-3 mol of OH- will convert 2.00 x 10-3 mol of CH3COOH

to CH3COO-.

initial

addition

final

CH3COOH (aq)

20.0 x 10-3 mol

18.0 x 10-3 mol

OH- (l)

?

0

2.00 x 10-3 mol

0

+

CH3COO- (aq) + H2O(l)

0

2.00 x 10-3 mol

new [CH3COOH] = n/Vf = (18.0 x 10-3 mol)/(60.0 mL)(10-3 L/mL) = 0.300 M

new [CH3COO-] = n/Vf = (2.00 x 10-3 mol)/(60.0 mL)(10-3 L/mL) = 0.033 M

pH = pKa + log

[base]

[acid]

= 4.74 + log

[0.033]

= 4.74 - 0.357 =

[0.300]

3.78

OR

[I]

[C]

[E]

CH3COOH (aq) + H2O(l) ?

0.300 M

-x

(0.300 -x)

CH3COO- (aq) + H3O+ (aq)

0.033 M

0

+x

+x

(0.033 + x)

x

Ka = (0.033 + x)x/(0.300 - x) ¡Ö 0.033x/0.300

¡à x = [H3O+] = 1.636 x 10-4

¡à pH = 3.79

(same within

rounding errors)

(c) (3 points) You don¡¯t have to add acid or base to a solution to change the

pH, you can instead dilute the solution. What is the new pH if 100 mL of a

solution of HCl with pH = 0.52 is diluted by addition of 200 mL of pure

water?

original [H3O+] = antilog(-pH) = 0.30 M

Use MiVi = MfVf where Vf = 100 mL + 200 mL = 300 mL

¡à Mf = (0.30 M)(100 mL)/(300 mL) = 0.10 M

¡à new pH = -log(0.10)

¡à pH = 1.00

4.

(5 points) Calculate the solubility of the salt M3X4 (containing M4+ and X3ions) if the Ksp = 4.2 x 10-8

I

C

E

M3X4 (s) ? 3 M4+ (aq) +

(solid)

0

-x

+3x

(solid)

3x

Ksp = 4.2 x 10-8 = [M4+]3[X3-]4

¡à Ksp = (3x)3(4x)4

¡à 4.2 x 10-8 = 6912 x7

x7 = 6.076 x 10-12

¡à x = 0.025 M

solubility of M3X4 = 0.025 M

4 X3- (aq)

0

+4x

4x

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download