Question

Kb for ammonia (NH3) is 1.76 x 10-5 What is Ka for ammonium, its conjugate acid. (NH4+) 3 sig figs Answer: 5.68x10^-10 Using the answer above, what is the pH of a 0.140 M solution of NH4+?

Answer 1

The Ka for the conjugate acid of ammonia (NH3), ammonium (NH4+), is 5.6x10^-10. To find the pH of a 0.140 M solution of NH4+, we can use the equilibrium expression for the dissociation of NH4+. Using the expression for Ka and assumptions, we can set up an equation to calculate the concentration of H3O+ ions, and then convert it to pH.

Step-by-step explanation:

The Ka for the conjugate acid of ammonia (NH3), ammonium (NH4+), can be calculated using the equation Ka = Kw / Kb. Substituting the given values, the Ka for ammonium is 5.6x10^-10. To find the pH of a 0.140 M solution of NH4+, we need to consider that ammonium is a weak acid. We can use the expression for the acid dissociation constant (Ka) to calculate the concentration of H3O+ ions, and then convert it to pH.

First, we can set up the equilibrium expression for the dissociation of NH4+:

• NH4+ (aq) + H2O (l) ⇌ H3O+ (aq) + NH3 (aq)

Since NH4+ is a weak acid, we can assume that the concentration of NH4+ that reacts with water to form H3O+ and NH3 is small compared to the initial concentration of NH4+. Therefore, we can assume that the equilibrium concentration of NH4+ is equal to the initial concentration, which is 0.140 M.

Using the expression for Ka and the assumption that the concentration of NH4+ equals its initial concentration, we can set up the equation:

• Ka = [H3O+][NH3] / [NH4+]
• 5.6x10^-10 = [H3O+][NH3] / 0.140

We can rearrange the equation to solve for [H3O+]:

• [H3O+] = (5.6x10^-10)(0.140) / [NH3]

Finally, we can use the concentration of [H3O+] to calculate the pH using the equation pH = -log[H3O+].

Answer 2

The pH of the solution is 6.1.

What is a conjugate acid?

One proton is added to the original base by the conjugate acid, which can then donate that extra proton in a reverse reaction to reform the original base.

Similar to this, an acid's conjugate base can take on a proton to reform the original acid even though it has one fewer proton than the original acid did.

We know that;

Kb. Ka =
`1 * 10^{14`

Ka =
`1 * 10^{-14`/
`1.76 * 10^{-5`

Ka =
`5.68 * 10^{-10`

The equation is;

`NH_4^+ + H_2O ---- > H_3O^+ + NH3`

Ka = [
`H_3O^+`] [
`NH_3`]/
`NH_4^+`

If [
`H_3O^+`] = [
`NH_3`] = x

`5.68 * 10^{-10`=
`x^2`/0.14 - x

`5.68 * 10^{-10`(0.14 - x) =
`x^2`

7.95 *
`10^{-11` -
`5.68 * 10^{-10`x =
`x^2`

`x^2` +
`5.68 * 10^{-10`x - 7.95 *
`10^{-11` = 0

x = 0.00000089 M

pH = -log(0.00000089 M)

pH = 6.1