Question

I NEED HELP PLEASE, THANKS!

Answer 1

Answer:

Here's what I get.

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

$\rm K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.335 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.335}{1.8 \times 10^{-5}} = 28 000 > 400\\\\x \ll 0.335$

(b) Solve for [OH⁻]

$\dfrac{x^{2}}{0.335} = 1.8 \times 10^{-5}\\\\x^{2} = 0.335 \times 1.8 \times 10^{-5}\\x^{2} = 6.03 \times 10^{-6}\\x = \sqrt{6.03 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.46 \times 10^{-3}} \textbf{ mol/L}$

(c) Calculate the pOH

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

(d) Calculate the pH

pH = 14.00 - pOH = 14.00 - 2.61 = 11.39