Question

I NEED HELP PLEASE, THANKS!

Ammonia (NH3) is an example of a Brønsted-Lowry Base.
-Define the Brønsted-Lowry acid-base theory.
-What is the pH of an ammonia solution that has a concentration of 0.335 M? The Kb of ammonia is 1.8 × 10^–5.

Answer 1

Here's what I get.

Step-by-step explanation:

1. Brønsted-Lowry theory

An acid is a substance that can donate a proton to another substance.

A base is a substance that can accept a proton from another substance.

2. pH of ammonia

The chemical equation is

`\rm NH$_(3)$ + \text{H}$_(2)$O \, \rightleftharpoons \,$ NH$_(4)^(+)$ + \text{OH}$^(-)$`

For simplicity, let's re-write this as

`\rm B + H$_(2)$O \, \rightleftharpoons\,$ BH$^(+)$ + OH$^(-)$`

(a) Set up an ICE table.

B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹: 0.335 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.335 + x x x

`\rm K_{\text{b}} = \frac{\text{[BH}^(+)]\text{[OH}^(-)]}{\text{[B]}} = 1.8 * 10^(-5)\\\\(x^(2))/(0.335 - x) = 1.8 * 10^(-5)`

Check for negligibility:

`(0.335)/(1.8 * 10^(-5)) = 28 000 > 400\\\\x \ll 0.335`

(b) Solve for [OH⁻]

`(x^(2))/(0.335) = 1.8 * 10^(-5)\\\\x^(2) = 0.335 * 1.8 * 10^(-5)\\x^(2) = 6.03 * 10^(-6)\\x = \sqrt{6.03 * 10^(-6)}\\x = \text{[OH]}^(-) = \mathbf{2.46 * 10^(-3)} \textbf{ mol/L}`

(c) Calculate the pOH

`\text{pOH} = -\log \text{[OH}^(-)] = -\log(2.46 * 10^(-3)) = 2.61`

(d) Calculate the pH

pH = 14.00 - pOH = 14.00 - 2.61 = 11.39