Calculate the [H+], [OH ‾ ], and the pH of 0.16 M solutions of:
hydroxylamine (HONH2, Kb = 1.1 10-8)
HONH2 I'll call RNH2
..........RNH2 + HOH ==> RNH3^+ + OH^-
initial...0.16M............0.......0
change......-x..............x......x
equil.....0.16-x............x......x
Kb = (RNH2)(OH^-)/RNH2
Substitute from the ICE chart above and solve for x which = OH^-, then convert to H^+ and pH.
To calculate the [H+], [OH-], and pH of a solution of hydroxylamine (HONH2), we need to use the concept of the dissociation of the compound and the equilibrium constant expression.
The dissociation of hydroxylamine in water can be represented as follows:
HONH2 + H2O ⇌ H3O+ + ONH2-
The equilibrium constant, Kb, is given as 1.1 x 10^-8. Kb represents the equilibrium constant for the reaction when hydroxylamine reacts with water to form the hydroxylammonium ion (H3O+) and the hydroxylamine ion (ONH2-).
Kb = [H3O+][ONH2-]/[HONH2]
Since we are given the concentration of the HONH2 solution as 0.16 M, we can assume that the concentration of HONH2 is equal to its initial concentration.
[HONH2] = 0.16 M
Now, we need to set up an ICE table to represent the initial, change, and equilibrium concentrations of the species involved in the reaction:
HONH2 + H2O ⇌ H3O+ + ONH2-
Initial: 0.16 M 0M 0M 0M
Change: -x -x +x +x
Equilibrium: 0.16 - x 0 - x x x
Since x is very small compared to 0.16, we can neglect it in the (0.16 - x) term.
Now, we can use the equilibrium concentrations to write the equilibrium constant expression in terms of these concentrations:
Kb = [H3O+][ONH2-]/[HONH2] = x * x / (0.16 - x)
The Kb value is given as 1.1 x 10^-8. We can substitute this value into the equation:
1.1 x 10^-8 = x^2 / (0.16 - x)
Since x is very small compared to 0.16, we can neglect it in the denominator.
1.1 x 10^-8 = x^2 / 0.16
Now we can solve for x:
x^2 = (1.1 x 10^-8) * 0.16
x^2 = 1.76 x 10^-9
x ≈ 1.33 x 10^-5
Now that we have the concentration of H3O+ (x), we can calculate the concentration of OH- as follows:
[H3O+] = [OH-] = 1.33 x 10^-5 M
Using the equation:
pH = -log[H+]
We can calculate the pH:
pH = -log(1.33 x 10^-5) ≈ 4.88
Therefore, the [H+], [OH-], and pH of the 0.16 M hydroxylamine solution are: [H+] = [OH-] = 1.33 x 10^-5 M and pH = 4.88.