Can somone please explain this problem to me? The hydrolysis equilibrium constant (Ka) for NH2NH3+ is 5.88E-9 (eq. 1).
(eq. 1): NH2NH3+(aq) + H2O(l) = H3O+(aq) + NH2NH2(aq)
Calculate the value of the base dissociation constant (Kb) for NH2NH2 as shown in the reaction (eq. 2).
(eq. 2): NH2NH2(aq) + H2O(l) = OH-(aq) + NH2NH3+(aq)
I solved it but I was one decimal place off. I am double checking but I cannot find my source of error. My answer was 1.7006E-5 by the way.
Kb = Kw/Ka = 1 x 10^-14/5.88 x 10^-9 = 1.7 x 10^-6.
To calculate the value of the base dissociation constant (Kb) for NH2NH2 in equation (eq. 2), you can use the relationship between Ka and Kb.
The relationship between Ka and Kb for a conjugate acid-base pair is given by the equation:
Ka x Kb = Kw
Where Kw is the ion product of water, which is equal to 1.0 x 10^-14 at 25°C.
In equation (eq. 1), NH2NH3+ acts as an acid and donates a proton (H+) to water, forming H3O+ and NH2NH2. So NH2NH3+ is the conjugate acid of NH2NH2.
In equation (eq. 2), NH2NH2 acts as a base and accepts a proton (H+) from water, forming OH- and NH2NH3+. So NH2NH2 is the conjugate base of NH2NH3+.
Now, we can use the relationship between Ka and Kb to solve for Kb.
Given Ka = 5.88 x 10^-9 (eq. 1)
We can rearrange the equation to solve for Kb:
Kb = Kw / Ka
Substituting the values:
Kb = (1.0 x 10^-14) / (5.88 x 10^-9)
Now, let's calculate the value of Kb:
Kb = 1.7 x 10^-6
So, the correct value of the base dissociation constant (Kb) for NH2NH2 is 1.7 x 10^-6, not 1.7006 x 10^-5 as you calculated.
Make sure you correctly entered the values and performed the calculations without any rounding errors to obtain the correct answer.