Question

1. Calculate the pH of a solution of 0.250M lysine dihydrochloride (H3K2+)

2. pH of a solution 0.0895 M lysine hydrochloride (H2K+)

3. pH of a solution of 0.100 M lysine (HK)

4. pH of a solution of 0.168 M sodium salt of lysine (NaK)

Answer 1

We note down the pK_a values of lysine as

pK_a1 = 2.18 ===> K_a1 = 10^-2.18 = 6.61*10^-3

pK_a2 = 8.95 ===> K_a2 = 10^-8.95 = 1.12*10^-9

pK_a3 = 10.53 ===> K_a3 = 10^-10.53 = 2.95*10^-11

1. The equilibrium reaction (along with the ICE chart) is (initial concentration of H₃K²⁺ is 0.250 M)

H₃K²⁺ (aq) <====> H⁺ (aq) + H₂K⁺ (aq)

initial                             0.250                          0               0

change                            - x                              x               x

equilibrium               (0.250 – x)                    x                x     

Therefore, K_a3 = [H⁺][H₂K⁺]/[H₃K²⁺] = (x)(x)/(0.250 – x)

or, 6.61*10^-3 = x²/(0.250 – x)

or, 1.6525*10^-3 – 6.61*10^-3x = x²

or, x² + 6.61*10^-3x – 1.6525*10^-3 = 0

or, x = [-(6.61*10^-3) + {(6.61*10^-3)² – 4.(1).(- 1.6525*10^-3)}^1/2]2.(1)

or, x = [-(6.61*10^-3) + 0.0815]/2 = 0.0374

Therefore, [H⁺] = 0.0374 M and pH = -log₁₀[H⁺] = -log₁₀(0.0374 ) = 1.426 ≈1.43

Ans: The pH of 0.250 M solution of H₃K²⁺ is 1.43.

2. Again, we shall employ the ICE chart as before.

H₂K⁺ (aq) <=====> H⁺ (aq) + HK (aq)

initial                             0.0895                          0                0

change                              - x₁                             x₁               x₁

equilibrium              (0.0895 – x₁)                     x₁              x₁

As before, K_a2 = x₁²/(0.0895 – x₁)

or, 1.12*10^-9 = x₁²/(0.0895 – x₁)

or, 1.00*10^-10 – 1.12*10^-9x₁ = x₁²

or, x₁² + 1.12*10^-9x₁ – 1.00*10^-10 = 0

Therefore, x₁ = [-(1.12*10^-9) + {(1.12*10^-9)² – 4.(1).(-1.00*10^-10)}^1/2]/2

or, x₁ = [-(1.12*10^-9) + 2.00*10^-5]/2 = 1.00*10^-5 (we neglect the first term as it is too small compared to the second).

Therefore, [H⁺] = 1.00*10^-5 M and pH = -log₁₀(1.00*10^-5) = 4.99

Ans: The pH of 0.0895 M solution of H₂K⁺ is 4.99.

3. We set up the ICE chart as

HK (aq) <=====> H⁺ (aq) + K^- (aq)

initial                                 0.100                        0              0

change                                - x₂                           x₂            x₂

equilibrium                  (0.100 – x₂)                   x₂            x₂

Therefore, K_a3 = x₂²/(0.100 – x₂)

or, 2.95*10^-11 = x₂²/(0.100 – x₂)

or, x₂² + 2.95*10^-11x₂ – 2.95*10^-12 = 0

or, x₂ = [-(2.95*10^-11) + {(2.95*10^-11)² – 4.(1).(2.95*10^-12)}^1/2]/2

or, x₂ = 1.72*10^-6

Therefore, [H⁺] = 1.72*10^-6 M and pH = 5.76

Ans: The pH is 5.76.