Hydrazine (NH2NH2) is a weak base that ionizes in water as follows.

NH2NH2(aq) + H2O(l) <==> HNH2NH2^+(aq) + OH^−(aq)

Calculate the concentration of OH^− ions in a 0.20 M solution of hydrazine, given that the base-dissociation constant (Kb) for hydrazine is 1.3 ✕ 10^−6. (Note that 0.20 M is known as the initial, formal, or total concentration of hydrazin.)

hydrazine = BH

......BH + H2O ==> BH2^+ + OH^-
I....0.2M...........0.......0
C.....-x............x.......x
E....0.2-x..........x.......x

Write the Kb expression, plug in the E line and solve for x = OH^-

Post your work if you get stuck.

5.7 x 10-12

To calculate the concentration of OH^- ions in a 0.20 M solution of hydrazine, we need to use the base dissociation constant (Kb) and the initial concentration of hydrazine.

The equation for the dissociation of hydrazine in water is:

NH2NH2(aq) + H2O(l) <==> HNH2NH2^+(aq) + OH^-(aq)

The base dissociation constant (Kb) is given as 1.3 × 10^(-6). This constant relates the concentrations of the products and reactants in the equilibrium expression:

Kb = [HNH2NH2+][OH^-] / [NH2NH2]

Given that the initial concentration of hydrazine is 0.20 M, we can assume that at equilibrium, the concentration of [NH2NH2] will decrease by some value, while the concentrations of [HNH2NH2+] and [OH^-] will increase by the same value.

Let's assume that the change in concentration for [HNH2NH2+] and [OH^-] is 'x'. Since the initial concentration of hydrazine is 0.20 M, the concentration of [NH2NH2] at equilibrium will be (0.20 - x) M.

Now, we can write the equation for Kb:

Kb = [HNH2NH2+][OH^-] / [NH2NH2]
1.3 × 10^(-6) = (x)(x) / (0.20 - x)

Solving this equation will give us the value of 'x', which represents the change in concentration for [HNH2NH2+] and [OH^-]. We can then use this value to calculate the concentration of OH^- ions.

I will solve this equation for you. Give me a moment to calculate the value of 'x'.

To calculate the concentration of OH⁻ ions in a 0.20 M solution of hydrazine, we can use the base-dissociation constant (Kb) and the stoichiometry of the reaction. Here's how you can do it:

Step 1: Write down the balanced chemical equation for the ionization of hydrazine:
NH2NH2(aq) + H2O(l) ⇌ HNH2NH2⁺(aq) + OH⁻(aq)

Step 2: Define the equilibrium concentration of OH⁻ ions as 'x'.

Step 3: Set up an expression for the base-dissociation constant (Kb):
Kb = ([HNH2NH2⁺][OH⁻]) / [NH2NH2]

Step 4: Since the concentration of OH⁻ ions is 'x' and the concentration of hydrazine is 0.20 M, the expression becomes:
Kb = (x)(x) / (0.20 - x)

Step 5: Substitute the value of Kb (1.3 × 10⁻⁶) into the equation:
1.3 × 10⁻⁶ = (x)(x) / (0.20 - x)

Step 6: Rearrange the equation and solve for 'x':
x² = (1.3 × 10⁻⁶)(0.20 - x)
x² = 2.6 × 10⁻⁷ - 1.3 × 10⁻⁶x
x² + 1.3 × 10⁻⁶x - 2.6 × 10⁻⁷ = 0

Step 7: Solve the quadratic equation using a suitable method, such as factoring, completing the square, or the quadratic formula. In this case, since the coefficients are small, we can factor it:
(x - 2.6 × 10⁻⁷)(x + 1.0 × 10⁻⁶) = 0

Step 8: Since the concentration cannot be negative, we discard the negative solution. Therefore:
x = 2.6 × 10⁻⁷ M

So, the concentration of OH⁻ ions in the 0.20 M solution of hydrazine is 2.6 × 10⁻⁷ M.