Alright, this time, I know I have not mistyped my answer.
Find the initial concentration of the weak acid or base in an aqueous solution of hydrazine (NH2NH2) with a pH of 10.16. Kb = 1.70 ×10−6 Answer in M
I've tried going from 10^-pH then using that (as my x) in the Kb = x^2/Initial then rearragning that to get x^2/Kb=Initial. I've even tried going from Ka, rearragning everything. But every time either way I get 0.0122899772M
Remember the system I use for online homework does not care about significant digits, it goes out to /at least/ 6 decimal places. And I butchered spelling rearranging.
I'll try to remember that but it isn't the usual. So all I can do is see if the set up is ok. Your set up on the earlier problem was ok.
That isn't your x in this hydrolysis equation. Calling NH2NH2 just BNH2, I would convert pH to pOH and use OH as x in the equation of
Kb = (x)(x)/(BNH2)
pH = 10.16; therefore, pOH = 14-10.16 = 3.84 and -3.84 = log(OH^-). I get OH^- = 1.44E-4M for OH^-
.........BNH2 + HOH ==> BNH3^+ + OH^-
I........y...............0........0
C........-x...............x........x
E.......y-x...............x........x
Kb = (x)(x)/(y-x) and solve for y.
To find the initial concentration of a weak base, hydrazine (NH2NH2), in an aqueous solution with a pH of 10.16, we need to use the Kb value and the equation for the reaction of the weak base with water:
NH2NH2 + H2O ⇌ NH3OH- + OH-
Here's the step-by-step process to find the answer:
1. Start by calculating the pOH of the solution. Since pH + pOH = 14, we can subtract the pH from 14:
pOH = 14 - 10.16
= 3.84
2. Convert the pOH back into OH- concentration using the formula 10^(-pOH):
[OH-] = 10^(-pOH)
= 10^(-3.84)
≈ 1.518 × 10^(-4) M
3. Since the base hydrazine, NH2NH2, reacts with water to form hydroxide ions (OH-), we can assume that the concentration of OH- is equal to the concentration of NH3OH-.
4. Utilize the Kb expression for the reaction NH2NH2 + H2O ⇌ NH3OH- + OH-:
Kb = [NH3OH-][OH-] / [NH2NH2]
The Kb value is given as 1.70 × 10^(-6) M.
5. Let's assume the initial concentration of NH2NH2 is x M.
6. Plug in the known values into the Kb expression:
1.70 × 10^(-6) = (1.518 × 10^(-4)) * (1.518 × 10^(-4)) / x
Simplify the equation:
1.70 × 10^(-6) = (2.303 × 10^(-8)) / x
7. Rearrange the equation to isolate x, the initial concentration of NH2NH2:
x = (2.303 × 10^(-8)) / (1.70 × 10^(-6))
x ≈ 0.013541 M
Therefore, the initial concentration of the weak base hydrazine (NH2NH2) in the aqueous solution with a pH of 10.16 is approximately 0.013541 M.