Hydrazine, N2H4(aq), is used in the preparation of polymers, pharmaceuticals and rocket fuel. Hydrazine has alkaline properties similar to ammonia and will act as a weak base. Given that a 100 mL sample of 0.10 mol/L hydrazine solution has a pH of 10.55 at 25.0 oC, write the formula for the conjugate acid and calculate the Kaof the conjugate acid.
I understand that we have to write the equation for the conjugate acid but I don't understand how to find the values then. I'm hoping someone can help me using the ICE (Initial, Change, Equilibrium) table.
See your post above.
To find the formula for the conjugate acid and calculate the Ka value, you can use the ICE table and the equation for the reaction between the hydrazine and its conjugate acid. Here's how you can approach the problem step-by-step:
Step 1: Write the equation for the reaction between hydrazine (base) and its conjugate acid.
N2H4(aq) + H2O(l) ⇌ N2H5+(aq) + OH-(aq)
Step 2: Construct the ICE table:
Initial: N2H4(aq) + H2O(l) ⇌ N2H5+(aq) + OH-(aq)
0.10 mol/L 0 0
Change: -x +x +x
Equilibrium: 0.10 - x x x
Step 3: Write the expression for the equilibrium constant (Ka):
Ka = [N2H5+][OH-] / [N2H4]
Step 4: Use the pH value to find the concentration of hydroxide ion ([OH-]):
pOH = 14 - pH
pOH = 14 - 10.55
pOH = 3.45
[OH-] = 10^-pOH
[OH-] = 10^-3.45
[OH-] = 3.55 x 10^-4 mol/L
Step 5: Since hydrazine is a weak base, we can assume that the concentration of the hydrazine will not change significantly during the reaction. Therefore, [N2H4] can be approximated as the initial concentration, which is 0.10 mol/L.
Step 6: Substitute the values into the equilibrium expression:
Ka = [(3.55 x 10^-4)(x)] / (0.10 - x)
Step 7: Since we assume that x is much smaller than 0.10, we can approximate (0.10 - x) as 0.10.
Ka = [(3.55 x 10^-4)(x)] / 0.10
Step 8: Simplify the equation:
Ka = 3.55 x 10^-3 x
Step 9: Since we assume that x is small, we can neglect it and assume it to be negligible. This assumption is valid because the initial concentration of hydrazine is much greater than the concentration of the products.
Step 10: Therefore, the expression for the Ka of the conjugate acid is:
Ka ≈ 3.55 x 10^-3
Hence, the formula for the conjugate acid of hydrazine is N2H5+. The approximate value of the Ka for the conjugate acid is 3.55 x 10^-3.
To find the formula for the conjugate acid and calculate the Ka of the conjugate acid, we can use the ICE (Initial, Change, Equilibrium) table and the concept of the equilibrium constant expression.
1. Write the equation for the dissociation of hydrazine into its conjugate acid and its conjugate base:
N2H4(aq) + H2O(l) ⇌ NH3OH+(aq) + OH-(aq)
2. Set up the ICE table:
N2H4(aq) + H2O(l) ⇌ NH3OH+(aq) + OH-(aq)
Initial: 0.10 0 0
Change: -x +x +x
Equilibrium: 0.10 - x x x
Here, x represents the change in concentration of NH3OH+ and OH- ions.
3. Since hydrazine is a weak base, it will partially dissociate to form NH3OH+ and OH- ions. The OH- ions will increase the concentration of hydroxide ions in the solution, leading to the pH of 10.55. We can use the pH value to calculate the concentration of OH- ions.
pOH = 14 - pH
pOH = 14 - 10.55
pOH = 3.45
[OH-] = 10^(-pOH)
[OH-] = 10^(-3.45)
[OH-] = 3.55 x 10^(-4) mol/L
Since the concentration of OH- ions is the same as the concentration of NH3OH+ ions (as one NH3OH+ is formed for every OH- ion), we can substitute [OH-] as the equilibrium concentration of NH3OH+ in the ICE table.
4. Calculate the Ka of the conjugate acid.
Ka = [NH3OH+] [OH-] / [N2H4]
Ka = (x)(x) / (0.10 - x)
Since we know that [OH-] = x = 3.55 x 10^(-4) mol/L, we can substitute these values into the equation:
Ka = (3.55 x 10^(-4))(3.55 x 10^(-4)) / (0.10 - 3.55 x 10^(-4))
Simplifying the equation, we get:
Ka = 1.26 x 10^(-7) mol/L
Therefore, the formula for the conjugate acid is NH3OH+ and the Ka of the conjugate acid is 1.26 x 10^(-7) mol/L.