For the reaction of hydrazine (N2H4) in water, Kb is 3.0 10-6.

H2NNH2(aq) + H2O(l)--> H2NNH3+(aq) + OH-(aq)
Calculate the concentrations of all species and the pH of a 1.6 M solution of hydrazine in water.
[OH-] =
[H2NNH3+] =
[H2NNH2] =
[H+] =
pH =

Oh, hydrazine, that's a fancy chemical name! Let's dive right into it and calculate the concentrations, shall we?

First, we have a 1.6 M solution of hydrazine (N2H4) in water. Since hydrazine is a weak base, it will react with water to form its conjugate acid, H2NNH3+, and hydroxide ions (OH-).

To find the concentrations of all the species, we'll use the Kb value, which is given as 3.0 x 10^-6.

Since we're starting with a 1.6 M solution of hydrazine, the concentration of hydrazine ([H2NNH2]) is, you guessed it, 1.6 M.

Now, let's use a little algebra to find the concentrations of [OH-] and [H2NNH3+].

We can let x be the concentration of [OH-] and [H2NNH3+]. Remember that for every molecule of hydrazine that reacts, one molecule of hydroxide ion (OH-) is formed.

The balanced equation tells us that the reaction ratio is 1:1. Considering this, the concentration of [OH-] would also be x, and the concentration of [H2NNH3+] would be x as well.

Using the Kb expression, Kb = [H2NNH3+][OH-]/[H2NNH2], and plugging in the values, we get:

3.0 x 10^-6 = x^2 / 1.6

Solving for x, we find x = 6.75 x 10^-4.

Therefore, [OH-] = [H2NNH3+] = 6.75 x 10^-4 M.

Since the solution is neutral, we can assume that the concentration of [H+] is equal to the concentration of [OH-]. Therefore, [H+] = 6.75 x 10^-4 M.

Now, for the final part, finding the pH! The pH is defined as the negative logarithm of the concentration of [H+]. Using the concentration of [H+] we found earlier, the pH would be:

pH = -log(6.75 x 10^-4) ≈ 3.17

So, the pH of a 1.6 M solution of hydrazine in water is approximately 3.17.

Hope that puts a smile on your face!

To solve this problem, we need to consider the reaction between hydrazine and water and use the given Kb value to calculate the concentrations of all the species and the pH of the solution.

From the balanced equation:

H2NNH2(aq) + H2O(l) --> H2NNH3+(aq) + OH-(aq)

Let's denote the initial concentration of hydrazine as [H2NNH2]₀ = 1.6 M.
We can assume that initially no [H2NNH3+] and [OH-] are present, so their concentrations are zero.

First, let's calculate the concentration of OH- using the Kb value:

Kb = [H2NNH3+][OH-] / [H2NNH2]

Since [H2NNH3+] and [OH-] are initially zero and [H2NNH2] is our only reactant, at equilibrium, we can assume that the concentration of [H2NNH2] that reacts is approximately equal to the initial concentration of hydrazine.

Therefore, Kb = [H2NNH3+][OH-] / 1.6

Rearranging the equation to solve for [OH-]:

[OH-] = Kb * 1.6 / [H2NNH2]

[OH-] = (3.0 * 10^(-6)) * 1.6 / 1.6

[OH-] = 3.0 * 10^(-6)

Therefore, [OH-] = 3.0 * 10^(-6) M.

Now, to calculate [H2NNH3+], we can use the fact that [H2NNH3+] is equal to [OH-] since the reaction is 1:1:

[H2NNH3+] = [OH-] = 3.0 * 10^(-6) M.

For [H+], we can use the fact that the equilibrium constant for water (Kw) can be used to find [H-].

Kw = [H+][OH-]

Since Kw = 1.0 * 10^(-14), and [OH-] = 3.0 * 10^(-6) M, we can solve for [H+] as follows:

1.0 * 10^(-14) = [H+]*(3.0 * 10^(-6))

[H+] = (1.0 * 10^(-14))/(3.0 * 10^(-6))

[H+] = 3.33 * 10^(-9) M.

Finally, we can use the [H+] value to calculate the pH using the formula:

pH = -log[H+]

pH = -log(3.33 * 10^(-9))

pH = 8.48

Therefore, the concentrations of the species and pH of a 1.6 M solution of hydrazine in water are:

[OH-] = 3.0 * 10^(-6) M
[H2NNH3+] = 3.0 * 10^(-6) M
[H2NNH2] = 1.6 M
[H+] = 3.33 * 10^(-9) M
pH = 8.48

To calculate the concentrations of all species and the pH of a 1.6 M solution of hydrazine in water, we need to use the Kb (base dissociation constant) value to determine the concentrations of the species involved in the reaction.

Step 1: Write the balanced equation for the reaction and identify the species:

H2NNH2(aq) + H2O(l) → H2NNH3+(aq) + OH-(aq)

In this reaction, hydrazine (H2NNH2) acts as a base and reacts with water to form the conjugate acid H2NNH3+ and the hydroxide ion OH-.

Step 2: Set up an ICE table and define the initial, change, and equilibrium concentrations:

[H2NNH2] is the initial concentration of hydrazine, given as 1.6 M.
[H2NNH3+] and [OH-] are the equilibrium concentrations we need to find.
[H2O] does not change significantly, so we can treat it as a constant and ignore it for the purpose of this calculation.

Species Initial (M) Change (M) Equilibrium (M)
----------------------------------------------------
H2NNH2 1.6 -x 1.6 - x
H2NNH3+ 0 +x x
OH- 0 +x x

Step 3: Write the expression for the Kb (base dissociation constant):

Kb = [H2NNH3+][OH-] / [H2NNH2]

Step 4: Substitute the relevant concentrations into the Kb expression:

Kb = (x)(x) / (1.6 - x) = 3.0 x 10^-6

Step 5: Solve for x:

Since [H2NNH3+] = [OH-] = x, we can rewrite the expression:

( x )^2 / (1.6 - x ) = 3.0 x 10^-6

Simplifying the equation leads to a quadratic equation:

x^2 = (3.0 x 10^-6)(1.6 - x)

Solving this quadratic equation will give us the concentration x of both [H2NNH3+] and [OH-].

Step 6: Calculate x and the concentrations of all species:

After solving the quadratic equation, let's assume x = [H2NNH3+] = [OH-].

Step 7: Calculate [H2NNH2] and [H+] using the concentrations determined earlier:

[H2NNH2] = 1.6 - x
[H+] = (1.0 x 10^-14) / x

Step 8: Calculate pH:

pH = -log[H+]

By plugging the calculated [H+] value into the equation, you can find the pH of the solution.

Please note that the calculation steps might be lengthy and require solving a quadratic equation. It is recommended to use a calculator or a software tool to obtain the final numerical values.

.....H2NNH2 + HOH ==> H2NNH3^+ + OH^-

I.....1.6...............0.........0
C.....-x.................x........x
E....1.6-x..............x.........x

Kb = 3.0E-6 = (H2NNH2^+)(OH^-)/(H2NNH2)
Substitute the E line into the Kb expression and solve for x.
That will give you H2NNH3^+, H2NNH2, OH^-. From that you can calculate pOH = -log(OH^^-), then get pH from
pH + pOH = pKw = 14