How many grams of NH2OH are needed to prepare 250mL of a solution with a pH of 9.56?
kb=6.6x10^-9
To calculate the grams of NH2OH needed to prepare the solution, we need to use the equation for the base dissociation of NH2OH:
NH2OH + H2O ⇌ NH3OH+ + OH-
Since we are given the pH of the solution, we can calculate the concentration of OH- ions using the following equation:
pOH = 14 - pH
So in this case, pOH = 14 - 9.56 = 4.44.
Next, we need to calculate the concentration of OH- ions (which is equal to the concentration of NH3OH+) using the equation:
pOH = -log[OH-]
We can rearrange this equation to solve for [OH-]:
[OH-] = 10^(-pOH)
Substituting the value of pOH we obtained above, we get:
[OH-] = 10^(-4.44)
Now, we can calculate the concentration of NH2OH (base) using the equation for the base dissociation constant (Kb):
Kb = [NH3OH+][OH-] / [NH2OH]
We know that Kb = 6.6x10^(-9), [NH3OH+] is the same as [OH-], and [NH2OH] is the unknown concentration we want to calculate. Substituting these values into the equation:
6.6x10^(-9) = (10^(-4.44))(10^(-4.44)) / [NH2OH]
Simplifying the equation:
6.6x10^(-9) = 10^(-8.88) / [NH2OH]
Now, solve the equation for [NH2OH]:
[NH2OH] = 10^(-8.88) / 6.6x10^(-9)
[NH2OH] = (10^(-8.88)) x (10^9 / 6.6)
[NH2OH] = 10^(0.12) / 6.6
[NH2OH] = 1.12 / 6.6
[NH2OH] = 0.1697 mol/L
Finally, we need to convert the concentration to grams by using the molecular weight of NH2OH, which is approximately 33.04 g/mol.
Mass of NH2OH = concentration x volume x molecular weight
Mass of NH2OH = 0.1697 mol/L x 0.250 L x 33.04 g/mol
Mass of NH2OH = 1.4206 g
Therefore, approximately 1.42 grams of NH2OH are needed to prepare 250 mL of a solution with a pH of 9.56.