Calculate the \rm pH of a 0.10 M solution of hydrazine, \rm N_2H_4. K_b for hydrazine is 1.3\times 10^{-6}.
........N2H4 + HOH ==> N2H5^+ + OH^-
I.......0.1M...........0.........0
C........-x.............x........x
E......0.1-x............x........x
Substitute the E line into Kb expression and solve for x = OH^-, then convert to pH.
To calculate the pH of a solution of hydrazine, we need to consider the basicity of the compound.
Hydrazine, N2H4, is a weak base. Weak bases partially dissociate in water, releasing hydroxide ions (OH-) into the solution. The equilibrium expression for the dissociation of hydrazine in water is as follows:
N2H4 + H2O ⇌ N2H5+ + OH-
The Kb value for hydrazine is given as 1.3 × 10^-6. Kb is the equilibrium constant for the reaction between the base and water to form the conjugate acid and hydroxide ions.
To calculate the OH- concentration in the solution, we can use the Kb value as follows:
Kb = [N2H5+][OH-] / [N2H4]
Since we have a concentration of 0.10 M for hydrazine,
[N2H4] = 0.10 M
However, we assume that the concentration of [N2H5+] formed is very small compared to the initial concentration of hydrazine, so we can approximate it as negligible. Therefore, we have:
Kb = [OH-] * [N2H4]
[OH-] = Kb / [N2H4]
[OH-] = (1.3 × 10^-6) / 0.10
[OH-] = 1.3 × 10^-5 M
Now that we know the concentration of hydroxide ions, we can determine the concentration of hydrogen ions (H+) using the following relationship:
[H+] * [OH-] = 1.0 x 10^-14
[H+] = (1.0 x 10^-14) / [OH-]
[H+] = (1.0 x 10^-14) / (1.3 × 10^-5)
[H+] = 7.7 × 10^-10 M
Finally, we can calculate the pH, which is defined as the negative logarithm (base 10) of the hydrogen ion concentration:
pH = -log[H+]
pH = -log(7.7 × 10^-10)
pH ≈ 9.11
Therefore, the pH of a 0.10 M solution of hydrazine is approximately 9.11.