A) Calculate [OH^-] in a o.12 M solution of hydrazine, H2NNH2(aq), having K of 3.0x10^-6.
B) what is the percent ionization of the H2NNH2(aq)
I can't get these for the life of mee!
If you will show your work I'll look for the error. I think OH^- is about 6E-4 and %ionization = about 0.50%
What i got was :
a) K = (3.0x10^-6)(0.12)/(0.12)(0.12)= 0.6
b) i got 1.90%
The error is you divided by 0.12 twice; once is all that is needed AND you didn't take the square root.
N2H4 + HOH ==> N2H5^+ + OH^-
Kb = (N2H5^+)(OH^-)/(N2H4)
3.0E-6 = (x)(x)/(0.12)
x^2 = sqrt(3.0E-6*0.12)
I obtained 6.0E-4 for OH^-
Then %ion = [(OH^-)/(0.12)]*100 = ?
I obtained 0.50%
No worries! I'll help you break down the steps to solve these problems.
A) To calculate [OH^-] in a solution of hydrazine (H2NNH2(aq)), we need to use the given K value. The reaction involved is:
H2NNH2(aq) ⇌ 2NH2^-(aq)
The stoichiometry of the reaction shows that for every 1 mole of hydrazine that ionizes, 2 moles of OH^- ions are produced. Thus, [OH^-] would be twice the concentration of the hydrazine that ionizes.
1. Write the equilibrium expression using the given K value:
K = [NH2^-]^2 / [H2NNH2]
2. Substitute the given concentration of hydrazine (0.12 M) into the equilibrium expression:
0.12 = (2x)^2 / (0.12 - x)
Here, x represents the amount of hydrazine that ionizes. Since x will be very small compared to 0.12 (because it's a weak electrolyte), we can approximate (0.12 - x) as just 0.12.
3. Rearrange and solve the equation for x:
0.12 = (2x)^2 / 0.12
0.12 * 0.12 = (2x)^2
0.0144 = 4x^2
x^2 = 0.0144 / 4
x^2 = 0.0036
x ≈ 0.06
So, the approximate concentration of hydrazine that ionizes is 0.06 M. Since [OH^-] is twice this value, [OH^-] ≈ 2 * 0.06 = 0.12 M.
B) The percent ionization can be determined by comparing the concentration of hydrazine that ionized (0.06 M) to the initial concentration of hydrazine (0.12 M) and expressing it as a percentage:
Percent Ionization = (0.06 M / 0.12 M) * 100%
Percent Ionization = 50%
Therefore, the percent ionization of H2NNH2(aq) is 50%.