Acid-Base Equilibria

A-salted

100

Determine the acetic acid concentration in a solution with [CH₃CO₂⁻] = 0.050 M and [OH⁻] = 2.5 × 10⁻⁶ M at equilibrium. 

C₂H₃O₂⁻(aq) + H₂O(l) ⇌ HC₂H₃O₂(aq) + OH⁻(aq)

K_a for HCH₃CO₂ = 1.8 x 10^-5

Kb = Kw/Ka = 5.6 x 10^-10

[HCH₃CO₂](2.5 × 10⁻⁶)/(0.050) = 5.6 x 10^-10

[HCH₃CO₂] = 1.1 x 10^-5 M

200

What is the pH of a 0.350 M solution of NaHCO₃, given that the K_b of HCO₃^- is 

2.30 × 10⁻⁸?

[H⁺] = x

x = (0.350 x 2.3 × 10⁻⁸)^1/2 = 9.0 x 10^-5

pOH = -log(9.0 x 10^-5) = 4.05

pH = 14.00 - 4.05 = 9.95

300

Calculate the pH of a solution that is 0.100 M in KCN. K_a for HCN = 2.1 x 10^-9

R     CN^-     +      H2O      ⇌      HCN      +      OH^-

I   0.100M           N/A                0                  ~0

C      -x                                    +x                 +x

E    0.100-x                                x                   x

Kb = Kw/Ka = 4.76 x 10^-6

4.76 x 10^-6 = x²/(0.100 - x)

x = 6.90 x 10^-4

pOH = -log(6.90 x 10^-4) = 3.16

pH = 14.00 - 3.16 = 10.84

400

Calculate the pH of a 1.20 M  NH₄NO₃ solution. K_b for NH₃ = 1.8 x 10^-5

R       NH₄⁺     +       H₂O       ⇌      NH₄OH      +      H⁺

I        1.2M              N/A                     0                 ~0

C        -x                                            x                   x

E         1.2-x                                       x                   x

Ka = Kw/Kb = 1.0 x 10^-14/1.8 x 10^-5 = 5.6 x 10^-10

5.6 x 10^-10 = x²/(1.2-x) 

x = 2.59 x 10^-5

pH = -log (2.59 x 10^-5) = 4.59

500

Lithium cyanide, LiCN, is a salt formed from by the reaction of the weak acid hydrocyanic acid, HCN, and the strong base lithium hydroxide. The Ka of hydrocyanic acid is 4.90×10−10.

Using this information, determine the pH of a 0.119 M solution of LiCN that is prepared at 25∘C.

R        CN^-        +       H₂O   --->        HCN      +      OH^-

I        0.119             N/A                    0                 ~0

C          -x                                         x                   x

E       0.119 - x                                  x                   x

Kb = Kw/Ka = 2.04 x 10^-5

sqrt(2.04 x 10^-5 x 0.119) = 0.00156

pOH = -log(0.00156) = 2.81

pH = 11.19

Weak Acid/Base pH

100

The pH of a 0.0516 M solution of nitrous acid, HNO₂, is 2.34. What is its Ka?

*Nitric acid HNO₃ is not the same as nitrous acid, HNO₂

R    HNO₂      + H₂O      --->      H⁺        NO₂^-

I     0.0516       N/A                  ~0         0

C     -x                                     x           x

E    0.0516 - x                          x           x

pH = -log[H⁺] = 2.34

[H⁺] = 10^-2.34 = 0.0046 M 

[HNO₂]_eq = 0.0516 - 0.0046 = 0.0470

Ka = (0.0046)(0.0046)/(0.0470) = 4.5 x 10^-4

200

HCl                             >>1

HCN                         2.9 x 10^-9

HC₂H₃O₂                  1.8 x 10^-5       

H₂O                        1.8 x 10^-16

H₂S                         5.7 x 10^-8

Rank the acids in terms of increasing acidity (1 = least acidic, 5 = most acidic)

1. H2O

2. HCN

3. H₂S

4. HC₂H₃O₂

5. HCl

300

What is the concentration of hydrogen ion and the pOH in a 0.534-M solution of formic acid, HCO₂H?

Ka = 1.8 x 10^-4

9.8 x 10^-3 = sqrt(1.8 x 10^-4 x 0.534) = x = H⁺

pH = -log (9.8 x 10^-3) = 2.01 

pOH = 14.00 - 2.01 = 11.99

400

If the value for Ka is very large, what will be the magnitude of the Kb for the conjugate base?

Kb will be very small.