- Define the terms below:

- Bronsted-Lowry acid - a proton donor

- Bronsted-Lowry base - a proton acceptor

- Lewis acid - an electron pair acceptor

- Lewis base - an electron pair donor

- amphoteric - having both basic and acidic properties

- pH the -log[H
^{+}]

- pOH - the -log[OH
^{-}]

- Bronsted-Lowry acid - a proton donor
- What is the pH of an aqueous solution that is 0.00897 M in hydrogen ion (H
^{+}_{(aq)}) concentration? What is the pOH?

pH = -log [H^{+}] = -log(0.00897) = -(-2.05) = 2.05

pH + pOH = 14 so pOH = 14 -pH = 14 -2.05 =11.95

- What is the hydroxide ion concentration ( [OH
^{-}] ) of an aqueous solution that has a pH of 13.31? What is the hydrogen ion concentration ( [H^{+}] )?

pH = -log[H^{+}] so [H^{+}] = 10^{-pH}= 10^{-13.31}= 4.90 x 10^{-14}

pOH = 14 - pH = 14 - 13.31 = 0.69

pOH = -log[OH^{-}] so [OH^{-}] = 10^{-pOH}= 10^{-0.69}= 0.204 M

- Put the acids in each part below in order from strongest to weakest.)

- HClO, HClO
_{4}, HClO_{2}, HClO_{3}

HClO_{4}> HClO_{3}> HClO_{2}> HClO

- H
_{3}SbO_{4}, H_{3}PO_{4}, H_{3}BiO_{4}, H_{3}AsO_{4}

H_{3}PO_{4}>H_{3}AsO_{4}>H_{3}SbO_{4}>H_{3}BiO_{4}

- HClO, HClO
- Find the pOH and pH of the solutions below:

- a solution where [H
^{+}] = 3.86 x 10^{-16}

pH = 15.41, pOH = -1.41

- a solution where [OH
^{-}] = 6.22 x 10^{-6}

pOH = 5.21, pH = 8.79

- a solution where [OH
^{-}] = [H^{+}]

pH = 7, pOH = 7

- a solution where the hydroxide ion concentration is equal to double the hydrogen ion concentration

The product of the concentrations must equal 1 x 10^{-14}

(2x)x = 1 x 10^{-14}= 2x^{2}

x^{2}= 5 x 10^{-15 }

x = (5 x 10^{-15})^{1/2}= 7.07 x 10^{-8}M = [H^{+}] pH = 7.15, pOH = 6.85

- a solution where [OH
^{-}] = 1.06 x 10^{-12}

pOH = 11.97, pH = 2.03

- a solution where [H
- What is the pH of a solution that is 4.850 M in NaNO
_{2}and 0.750 M in HNO_{2}? The K_{A}for HNO_{2}is 4.5 x 10^{-4}.

HNO_{2(aq)}º H^{+}_{(aq)}+ NO_{2}^{-}_{(aq)}

Ka[HNO_{2}]}/[NO_{2}^{-}] = {(4.5 x 10^{-4})(0.750)}/(4.850) = [H^{+}]

[H^{+}] = 6.96 x 10^{-5}

pH = -log[H^{+}] = - log(6.96 x 10^{-5}) = 4.16

- What is the pH of a buffer solution that is 2.00 M in sodium benzoate (NaOOCC
_{6}H_{5}) and 0.100 M in benzoic acid (HOOCC_{6}H_{5})? The K_{A}of benzoic acid is 6.3 x 10^{-5}.

HOOCC_{6}H_{5(aq)}<-----------> H^{+}_{(aq)}+^{-}OOCC_{6}H_{(aq)}

K_{a}= {[H^{+}][^{-}OOCC_{6}H_{5}]}/[HOOCC_{6}H_{5}]

{Ka[HOOCC_{6}H_{5}]}/[^{-}OOCC_{6}H_{5}] = [H^{+}]

{(6.3 x 10^{-5})(0.100)}/(2.00) = 3.15 x 10^{-6}= [H^{+}]

pH = -log[H^{+}] = -log(3.15 x 10^{-6}) = 5.50

- Using the buffer solution in the problem above, what is the pH if 15.00 mL of 2.50 M HCl
_{(aq)}is added to 75.00 mL of the buffer solution? What would be the pH if 25.00 mL of 1.45 M NaOH was added to 85.00 mL of the buffer solution in problem the problem above?

Using the acid solution:

Dilution problems:

[HOOCC_{6}H_{5}] = {(0.100 M)(75.00 mL)}/(90.00 mL) = 0.0833 M

[^{-}OOCC_{6}H_{5}] = {(2.00 M)(75.00 mL)}/(90.00 mL) = 1.67 M

Added acid:[H^{+}] = {(2.50 M)(15.00 mL)}/(90.00 mL) = 0.417 M

The added acid will decrease the concentration of the benzoate ion,^{-}OOCC_{6}H_{5},

new [^{-}OOCC_{6}H_{5}] = 1.67 - 0.417 = 1.253 M

The added acid will increase the benzoic acid, HOOCC_{6}H_{5},

new [HOOCC_{6}H_{5}] = 0.0833 + 0.417 = 0.500 M

Rearranging K_{a}= {[H^{+}][^{-}OOCC_{6}H_{5}]}/[HOOCC_{6}H_{5}] to solve for [H^{+}] yields [H^{+}] = {K_{a}[HOOCC_{6}H_{5}]}/[^{-}OOCC_{6}H_{5}]

[H^{+}] = {(6.3 x 10^{-5})(0.500)}/(1.253) = 2.61 x 10^{-5}

pH = -log[H^{+}] = -log(2.61 x 10^{-5}) = 4.58

For added base:

Dilution problems:

[^{-}OOCC_{6}H_{5}] = {(2.00 M)(85.00 mL)}/(110.00 mL) = 1.545 M

[HOOCC_{6}H_{5}] = {(0.100 M)(85.00 mL)}/(110.00 mL) = 0.0773 M

added base:

[OH^{-}] = {(1.45 M)(25.00 mL)}/(110.00 mL) = 0.330 M

The added base will decrease the acid concentration,

0.0773 M - 0.330 M = -0.2527 M

The negative number indicates that this buffer does not have enough capacity to handle this amount of base, leaving an excess of 0.2527 M in hydroxide. Therefore the pOH of the solution is:

pOH = -log[OH^{-}] = -log(0.2527) = 5.97

pH = 14 - pOH = 14 - 5.97 = 8.03

- What is the percent ionization for the following solutions of acids?

- 2.50 M HC
_{2}H_{3}O_{2, }K_{A}= 1.8 x 10^{-5}

HC_{2}H_{3}O_{2(aq)}º H^{+}_{(aq)}+ C_{2}H_{3}O_{2}^{-}_{(aq)}

K_{A}= {[H^{+}][C_{2}H_{3}O_{2}^{-}]}/[HC_{2}H_{3}O_{2}] = x^{2}/[HC_{2}H_{3}O_{2}]

K_{A}[HC_{2}H_{3}O_{2}] = x^{2}= (1.8 x 10^{-5})(2.50) = 4.5 x 10^{-5}

x = (4.5 x 10^{-5})^{1/2}= 6.71 x 10^{-3}= [H^{+}] = [C_{2}H_{3}O_{2}^{-}]

% ionization = {(6.71 x 10^{-3})/2.50} x 100 = 0.268%

quadratic not needed

- 0.00500 M HCOOH, K
_{A}= 1.7 x 10^{-4}

18.4%, quadratic needed

- 0.1500 M C
_{6}H_{5}OH, K_{A}= 1.3 x 10^{-10}

0.00294%, quadratic not needed

- 0.000515 M HF
_{(aq)}, K_{A}= 7.1 x 10^{-4}

completely ionized!! might be small difference with quadratic

- 0.00180 M HC
_{9}H_{7}O_{4}, K_{A}= 3.0 x 10^{-4}

40.8%, quadratic needed

- 2.50 M HC
- What is the sulfate concentration in a 14.00 M H
_{2}SO_{4(aq)}solution? K_{a1}= extremely large, ~100 % dissociation, K_{A2}= 1.3 x 10^{-2}.

H_{2}SO_{4(aq)}6 H^{+}_{(aq)}+ HSO_{4}^{-}_{(aq)}, [H^{+}] = [HSO_{4}^{-}] = 14.00 M

HSO_{4}^{-}_{(aq)}º H^{+}_{(aq)}+ SO_{4}^{2-}_{(aq)}

K_{A2}= {[H+][SO_{4}^{2-}]}/[HSO_{4}^{-}] but [H^{+}] = [HSO_{4}^{-}], since second dissociation provides an insignificant amount of H^{+}compared to first dissociation thus K_{A2}= [SO_{4}^{2-}] = 1.3 x 10^{-2}.

- What is the arsenate concentration (AsO
_{4}^{3-}) in 3.00 M arsenic acid (H_{3}AsO_{4})? K_{A1}= 6.0 x 10^{-3}, K_{A2}= 1.05 x 10^{-7}, K_{A3}= 3.0 x 10^{-12}

H_{3}AsO_{4(aq)}º H^{+}_{(aq)}+ H_{2}AsO_{4}^{-}_{(aq)}

% ionization calculation:

K_{A1}= {[H^{+}][H_{2}SO_{4}^{-}]}/[H_{3}AsO_{4}] = x^{2}/[H_{3}AsO_{4}]

K_{A1}[H_{3}AsO_{4}] = x^{2}

(6.0 x 10^{-3})(3.00) = 1.80 x 10^{-2}= x^{2}

(1.8 x 10^{-2})^{1/2}= x = 0.134

% ionization = {0.134/3.00} x 100 = 4.47%, quadratic not needed, thus [H^{+}] = [H_{2}AsO_{4}^{-}] = 0.134 M

H_{2}AsO_{4}^{-}_{(aq)}º H^{+}_{(aq)}+ HAsO_{4}^{2-}_{(aq)}

K_{A2}= {[H^{+}][HAsO_{4}^{2-}]}/[H_{2}AsO_{4}^{-}] but [H^{+}] = [H_{2}AsO_{4}^{-}] so K_{A2}= [HAsO_{4}^{2-}]

HAsO_{4}^{2-}_{(aq)}º H^{+}_{(aq)}+ AsO_{4}^{3-}_{(aq)}

K_{A3}= {[H^{+}][AsO_{4}^{3-}]}/[HAsO_{4}^{2-}] K_{A3}[HAsO_{4}^{2-}]/[H^{+}] = [AsO_{4}^{3-}] but [H^{+}] = 0.134 and [HAsO_{4}^{2-}] = K_{A2}

so K_{A3}K_{A2}/[H^{+}] = [AsO_{4}^{3-}]

{(3.0 x 10^{-12})(1.05 x 10^{-7})}/(0.134) = 2.35 x 10^{-18 }M = [AsO_{4}^{3-}]

- What is the pH of a solution that is

HYDROLYSIS!!

- 6.50 M in NaC
_{2}H_{3}O_{2}, HC_{2}H_{3}O_{2}K_{A}= 1.8 x 10^{-5}

C_{2}H_{3}O_{2}^{-}_{(aq)}+ H_{2}O º HC_{2}H_{3}O_{2(aq)}+ OH^{-}_{(aq)}

K_{B}= (1.0 x 10^{-14})/K_{A}= (1.0 x 10^{-14})/(1.8 x 10^{-5})

K_{B}= 5.56 x 10^{-10}

K_{B}= {[H_{2}C_{2}H_{3}O_{2}][OH^{-}]}/[C_{2}H_{3}O_{2}^{-}] = x^{2}/[C_{2}H_{3}O_{2}^{-}]

K_{B}[C_{2}H_{3}O_{2}^{-}] = x^{2}= (5.56 x 10^{-10})(6.50)

x^{2}= 3.614 x 10^{-9<}

x = (3.614 x 10^{-9})^{1/2}= 6.01 x 10^{-5}= [OH^{-}]

pOH = -log[OH^{-}] = -log(6.01 x 10^{-5}) = 4.22

pH = 14 -pOH = 14 -4.22 = 9.78

- 2.60 M in NaCN, HCN K
_{A}= 2.1 x 10^{-9}

same method as a, pH = 11.55

- 4.50 M in NH
_{4}Cl, NH_{3}K_{B}= 1.8 x 10^{-5}

NH_{4}^{+}_{(aq)}º H^{+}_{(aq)}+ NH_{3(aq)}

K_{A}= K_{w}/K_{B}= (1.0 x 10^{-14})/(1.8 x 10^{-5}) = 5.56 x 10^{-10}

K_{A}= {[H^{+}][NH_{3}]}/[NH_{4}^{+}] = x^{2}/[NH_{4}^{+}]

K_{A}[NH_{4}^{+}] = x^{2}= (5.56 x 10^{-10})(4.50) = 2.502 x 10^{-9}

x = (2.502 x 10^{-9})^{1/2}= 5.00 x 10^{-5 }= [H^{+}]

pH = -log[H^{+}] = -log(5.00 x 10^{-5}) = 4.30

- 2.34 M in NaIO
_{3}, HIO_{3}K_{A}= 1.58 x 10^{-1}

same method as a, pH = 7.59

- 2.95 M in NaN
_{3}, HN_{3}K_{A}= 1.9 x 10^{-5}

same method as a, pH = 9.60

- 6.50 M in NaC
- What is the pH of a buffer that is 3.93 M in NaCN and 0.951 M in HCN? If 10.00 mL of 0.1500 M HCl is added to 100.00 mL of this buffer what is the pH? If 15.00 mL of 0.6000 M NaOH is added to 95.00 mL of this buffer what is the pH?

HCN_{(aq)}º H^{+}_{(aq)}+ CN^{-}_{(aq)}K_{A}= 2.1 x 10^{-9}

K_{A}= {[H^{+}][CN^{-}]}/[HCN] = 2.1 x 10^{-9}

K_{A}[[HCN]/[CN^{-}] = [H^{+}] = {(2.1 x 10^{-9})(0.951)/(3.93)

[H^{+}] = 5.08 x 10^{-10}

pH = -log[H^{+}] = -log(5.08 x 10^{-10}) = 9.29

Added acid problem:

Dilutions:

[CN^{-}] = {(3.93 M)(100.00 mL)}/(110.00 mL) = 3.57 M

[HCN] = {(0.951 M)(100.00 mL)}/(110.00 mL) = 0.865 M

added acid:

[H^{+}] = {(0.1500 M)(10.00 mL)}/(110.00 mL) = 0.0136 M

Added acid will decrease [CN^{-}]

new [CN^{-}] = 3.57 - 0.0136 = 3.5564 M

Added acid will increase [HCN]

new [HCN] = 0.865 + 0.0136 = 0.8786 M

K_{A}[HCN]/[CN^{-}] = [H^{+}] = {(2.1 x 10^{-9})(0.8786)}/(3.5564)

[H^{+}] = 5.19 x 10^{-10}, pH = 9.28

Added base problem:

Dilutions:

[CN^{-}] = {(3.93 M)(95.00 mL)}/(110.00 mL) = 3.39 M

[HCN] = {(0.951 M)(95.00 mL)}/(110.00 mL) = 0.821 M

added base:

[OH^{-}] = {(0.6000 M)(15.00 mL)}/(110.00 mL) = 0.0818 M

Added base will decrease [HCN] and increase [CN^{-}]

new [CN^{-}] = 3.39 + 0.0818 = 3.4718 M

new [HCN] = 0.821 -0.0818 = 0.7392 M

_{A}[HCN]/[CN^{-}] = {(2.1 x 10^{-9})(0.7392)}/(3.4718) = [H^{+}]

[H^{+}] = 4.47 x 10^{-10}, pH = 9.35

HCN_{(aq)}º H^{+}_{(aq)}+ CN^{-}_{(aq) } - Design buffers with the at pH = 5.00 with the following

- a system 1.00 M in HC
_{2}H_{3}O_{2}, K_{A}= 1.8 x 10^{-5}

pH 5 means [H^{+}] = 1.00 x 10^{-5}

HC_{2}H_{3}O_{2(aq)}º H^{+}_{(aq)}+ C_{2}H_{3}O_{2}^{-}_{(aq)}

K_{A}= {[H^{+}][C_{2}H_{3}O_{2}^{-}]}/[HC_{2}H_{3}O_{2}]

K_{A}[HC_{2}H_{3}O_{2}]/[H^{+}] =

{(1.8 x 10^{-5})(1.00)}/(1.00 x 10^{-5})=[C_{2}H_{3}O_{2}^{-}] = 1.80 M

- a system 0.750 M in NaCN, HCN K
_{A}= 2.1 x 10^{-9}

[HCN] = 1.575 x 10^{-4}M

- a system 1.00 M in NH
_{3}, K_{B}= 1.8 x 10^{-5}

[NH_{4}^{+}] = 17985.6 M, a ridiculous answer, since the molarity of water in itself is the maximum at 55.55 M. You cannot make a buffer at this pH with these concentrations with this system .

- a system 1.00 M in NH
_{4}Cl, NH_{3}K_{B}= 1.8 x 10^{-5}

[NH_{3}] = 5.56 x 10^{-5}M. Although possible to make this buffer, it actually has almost no capacity.

- a system 1.00 M in NaC
_{2}H_{3}O_{2}, HC_{2}H_{3}O_{2}K_{A}= 1.8 x 10^{-5}

[HC_{2}H_{3}O_{2}] = 0.556 M

- a system 1.00 M in HC
- What are the equilibrium concentrations of the reactants and products below if a solution was made that was initially 1.00 M in Cl
_{3}CCOOH_{(aq)}, 0.500 M in H^{+}_{(aq) }and 0.325 M in Cl_{3}CCOO^{-}_{(aq)}? K_{A}= 1.29 x 10^{-1}

Cl_{3}CCOOH_{(aq)}º H^{+}_{(aq)}+ Cl_{3}CCOO^{-}_{(aq)}

K_{A}= {[H^{+}][Cl_{3}CCOO^{-}]}/[Cl_{3}COOH]

> K_{A}= {[0.500 + x][0.325 + x]}/[1.00 - x]

K_{A}[1.00 - x] = [0.500 + x][0.325 + x]

0.129 - 0.129x = x^{2}+ 0.825x + 0.1625

0 = x^{2}+ 0.954x + 0.0335

x = {-b__+__(b^{2}- 4ac)^{1/2}}/2a

x = {-0.954__+__[(0.954)^{2}-4(1)(0.0335)]^{1/2}}/2(1)

x = {-0.954__+__(0.910116 - 0.134)^{1/2}}/2

x = {-0.954__+__(0.776116)^{1/2}}/2

x = {-0.954__+__0.881}/2

x = -0.0365 or -0.9175

The only root that makes sense is the first since the second gives negative concentrations for [H^{+}] and [Cl_{3}COO^{-}]

Thus the equilibrium concentrations are:

[Cl_{3}COOH] = 1.00 - (-0.0365) = 1.0365__=__1.04 M

[Cl_{3}COO^{-}] = 0.325 + (-0.0365) = 0.2885__=__0.288 M

[H^{+}] = 0.500 + (-0.0365) = 0.4635__=__0.464 M

- What is the molar solubility of PbSO
_{4}if the K_{sp}for PbSO_{4}is 1.3 x 10^{-8}?

PbSO_{4(s)}º Pb^{2+}_{(aq)}+ SO_{4}^{2-}_{(aq)}, Ksp = [Pb^{2+}][SO_{4}^{2-}]

Since one mole yields 1 mole of Pb^{2+}and 1 mole of SO_{4}^{2-}Ksp = x^{2}therefore, x, the molar solubility is equal to

(Ksp)^{1/2}= (1.3 x 10^{-8})^{1/2}= 1.14 x 10^{-4}moles/L

- What is K
_{sp}of CaF_{2}if the molar solubility of CaF_{2}is 2.136 x 10^{-4}M?

CaF_{2(s)}º Ca^{2+}_{(aq)}+ 2 F^{-}_{(aq)}, K_{sp}= [Ca^{2+}][F^{-}]^{2}

therefore K_{sp}=(x)(2x)^{2}= 4x^{3}but x is equal to the molar solubility so K_{sp}= 4(2.136 x 10^{-4})^{3}= 3.898 x 10^{-11}

- If the molar solubility of Fe(OH)
_{2}is 5.85 x 10^{-6}, what is K_{sp}for Fe(OH)_{2}?

Fe(OH)_{2(s)}º Fe^{2+}_{(aq)}+ 2 OH^{-}_{(aq)}

K_{sp}= [Fe^{2+}][OH^{-}]^{2}

Using the same arguments as in the previsou problems, 5.85 x 10^{-6}moles of Fe^{2+}will be present, but twice as many moles (1.17 x 10^{-5}) of OH^{-}will be present.

Ksp = (5.85 x 10^{-6})(1.17 x 10^{-5})^{2}= 8.01 x 10^{-16}

- If the K
_{sp}for Mn(OH)_{2}is 2.00 x 10^{-13}, what is the molar solubility of Mn(OH)_{2}?

Mn(OH)_{2(s)}º Mn^{2+}_{(aq)}+ 2 OH^{-}_{(aq)}

For every Mn(OH)_{2}that dissociates one Mn^{2+}is formed but two OH^{-}s are formed thus the molar solubility may be calculated by using x for the Mn^{2+}concentration and 2x for the OH^{-}concentration. The quantity "x" will be equal to the molar solubility.

Ksp = x(2x)^{2}= 4x^{3}

Ksp/4 = x^{3}

(Ksp/4)^{1/3}= x

(2.00 x 10^{-13}/4)^{1/3}= x

(5.00 x 10^{-14})^{1/3}= x = 3.68 x 10^{-5}mol/L

- What is the molar solubility of AgCl in 1.00 M in NaCN?

Ksp_{AgCl}= 1.6 x 10^{-10}, Kf_{Ag(CN)2-}= 1.0 x 10^{21}

_{(s)}º Ag^{+}_{(aq)}+ Cl^{-}_{(aq)}

Ag^{+}_{(aq)}+ 2 CN^{-}_{(aq)}º Ag(CN)_{2}^{-}_{(aq)}

Overall rxn: AgCl_{(s)}+ 2 CN^{-}_{(aq)}º Ag(CN)_{2}^{-}_{(aq)}+ Cl^{-}_{(aq)}

K = KspK_{f}= {[Ag(CN)_{2}^{-}][Cl^{-}]}/[CN^{-}]^{2}

K = (1.6 x 10^{-10})(1.0 x 10^{21}) = 1.6 x 10^{11}

1.6 x 10^{11}=(s)(s)/(1.0-2s)^{2}

1.6 x 10^{11}= s^{2}/(1.0-2s)^{2}

(1.6 x 10^{11})^{1/2}= 4.0 x 10^{5}= s/(1.0 -2s)

4.0 x 10^{5}- (8.0 x 10^{5})s = s

4.0 x 10^{5}= (8.0 x 10^{5})s + s

s = 0.500 moles/L for molar solubility

- What is the molar solubility of Zn(OH)
_{2}? What is its molar solubility in a solution buffered at pH 14? At pH 5? The solubility product for Zn(OH)_{2}is 2.1 x 10^{-14}.

Zn(OH)_{2(s)}º Zn^{2+}_{(aq)}+ 2 OH^{-}_{(aq)}

K_{sp}= [Zn^{2+}][OH^{-}]^{2}

K_{sp}= x(2x)^{2}= 4x^{3}

Ksp/4 = 5.25 x 10^{-15}= x^{3}

x = (5.25 x 10^{-15})^{1/3}= 1.74 x 10^{-5}M = molar solubility

In a pH 14 buffer pOH = 14-14 = 0 so [OH^{-}] = 10^{0}= 1.00

so Ksp/[OH^{-}]^{2}= [Zn^{2+}] = (2.1 x 10^{-14})/(1.00)^{2}= 2.1 x 10^{-14}which is the molar solubility

In pH 5 buffer, pOH = 14 -5 = 9, [OH^{-}] = 1.00 x 10^{-9}

[Zn^{2+}] = Ksp/[OH^{-}]^{2}= (2.1 x 10^{-14})/(1.00 x 10^{-9})^{2}

[Zn^{2+}] = 2.1 x 10^{4}M (larger than maximum molarity, very soluble)

- Define:

- K
_{sp}- the solubility product equal to the reaction quotient of the equilibrium expression of a salt of limited aqueous solubility

- K
_{f}- formation constant for complex ions

- buffer - a solution that resist pH change

- buffer capacity - refers to how much added acid or base a buffer can absorb before coming non-functional

- K
_{d}dissociation constant for a complex ion, the inverse of K_{f}, i.e. 1/K_{f}

- common ion effect - refers to the shift in equilibrium that occurs when an ion common to the equilibrium is added (i.e. the effect of
LeChatelier's principle)

- pK
_{a}= -logK_{a}, which during a pH titration of an acid is equal to the pH at the half-equivalence point

- pK
_{b}= -lofK_{b}

- titration - a volumetric method for determining concentration of a material in solution

- molar soulubility - how many moles of a material can be dissolved in 1.00 L of water

- complex ion - an ion formed by the association of another species with an metal ion

(eg. Ag^{+}_{(aq)}+ 2 NH_{3(aq)}<-------> Ag(NH_{3})_{2}^{+}_{(aq)})

- K
- What is the solubility of AgCl in 5.00 M NH
_{3}?AgCl _{(s)}

K - s<---------> Ag ^{+}_{(aq)}

s+ Cl ^{-}_{(aq)}

sK _{sp}= 1.8 x 10^{-10}Ag ^{+}_{(aq)}

s+ 2 NH _{3(aq)}

5.00 - 2s<--------> Ag(NH _{3})_{2}^{+}_{(aq)}

sK _{f}= 1.7 x 10^{7}

K_{C}= K_{sp}K_{f}= (1.8 x 10^{-10})(1.7 x 10^{7}) = 3.06 x 10^{-3}

K_{C}= [Ag^{+}][Cl^{-}][Ag(NH_{3})_{2}^{+}]/[Ag^{+}][NH_{3}]^{2}

K_{C}= [Cl^{-}][Ag(NH_{3})_{2}^{+}]/[NH_{3}]^{2}

K_{C}= s^{2}/(5.00 - 2s)^{2}

(K_{C})^{1/2}= (s^{2}/(5.00 - 2s)^{2})^{1/2}= s/(5.00 - 2s) = (3.06 x 10^{-3})^{1/2}= 5.53 x 10^{-2}

s/(5.00 - 2s) = 5.53 x 10^{-2}

s = (5.00 -2s)(5.53 x 10^{-2}

s = 0.2765 - 0.1106s

s + 0.1106 s = 0.2765

1.1106s = 0.2765

s = 0.2765/1.1106 = 0.249 M

- Explain how the pK
_{A}of an acid may be determined with a pH titration

In general a weak acid may be reperesented by the equation:

HA _{(aq)}<--------> H^{+}_{(aq)}+ A^{-}_{(aq)}

thus K_{A}= [H^{+}][A^{-}]/[HA]

But at the half-equivalence point, half of the the HA has been consumed, and the concentrations of [HA] and [A^{-}] will be equal

thus at the half-equivalence point K_{A}= [H^{+}]

so at the half equivlance point -logK_{A}= -log[H^{+}]

and pK_{A}= pH

- Are the solutions below acidic or basic?

- 3.00 M Na
^{+}F^{-}- contains the conjugate base of the weak acid HF, solution will be basic

- 3.00 M Na
^{+}C_{2}H_{3}O_{2}^{-}- contains the conjugate base of the weak acid HC_{2}H_{3}O_{2}, solution will be basic

- 3.00 M NH
_{4}^{+}Cl- - contains the conjugate acid of the weak base NH _{3}, solution will be acidic

- 3.00 M CH
_{3}CH_{2}NH_{3}^{+}Cl^{-}- contains the conjugate acid of the weak base CH_{3}CH_{2}NH_{2}, solution will be acidic.

- 3.00 M Na
- What is the molar solubility of AgCl in 1.00 M in NaCN?

Ksp_{AgCl}= 1.6 x 10^{-10}, Kf_{Ag(CN)2-}= 1.0 x 10^{21}

_{(s)}º Ag^{+}_{(aq)}+ Cl^{-}_{(aq)}

Ag^{+}_{(aq)}+ 2 CN^{-}_{(aq)}º Ag(CN)_{2}^{-}_{(aq)}

Overall rxn: AgCl_{(s)}+ 2 CN^{-}_{(aq)}º Ag(CN)_{2}^{-}_{(aq)}+ Cl^{-}_{(aq)}

K = KspK_{f}= {[Ag(CN)_{2}^{-}][Cl^{-}]}/[CN^{-}]^{2}

K = (1.6 x 10^{-10})(1.0 x 10^{21}) = 1.6 x 10^{11}

1.6 x 10^{11}=(s)(s)/(1.0-2s)^{2}

1.6 x 10^{11}= s^{2}/(1.0-2s)^{2}

(1.6 x 10^{11})^{1/2}= 4.0 x 10^{5}= s/(1.0 -2s)

4.0 x 10^{5}- (8.0 x 10^{5})s = s

4.0 x 10^{5}= (8.0 x 10^{5})s + s

s = 0.500 moles/L for molar solubility

- What is the molar solubility of Zn(OH)
_{2}? What is its molar solubility in a solution buffered at pH 14? At pH 5? The solubility product for Zn(OH)_{2}is 2.1 x 10^{-14}.

Zn(OH)_{2(s)}º Zn^{2+}_{(aq)}+ 2 OH^{-}_{(aq)}

K_{sp}= [Zn^{2+}][OH^{-}]^{2}

K_{sp}= x(2x)^{2}= 4x^{3}

Ksp/4 = 5.25 x 10^{-15}= x^{3}

x = (5.25 x 10^{-15})^{1/3}= 1.74 x 10^{-5}M = molar solubility

In a pH 14 buffer pOH = 14-14 = 0 so [OH^{-}] = 10^{0}= 1.00

so Ksp/[OH^{-}]^{2}= [Zn^{2+}] = (2.1 x 10^{-14})/(1.00)^{2}= 2.1 x 10^{-14}which is the molar solubility

In pH 5 buffer, pOH = 14 -5 = 9, [OH^{-}] = 1.00 x 10^{-9}

[Zn^{2+}] = Ksp/[OH^{-}]^{2}= (2.1 x 10^{-14})/(1.00 x 10^{-9})^{2}

[Zn^{2+}] = 2.1 x 10^{4}M (larger than maximum molarity, very soluble)