Question

I need some help with a long chemistry problem. Anything is appreciated!

Answer 1

Answer:

The CSI is wrong.  

Explanation:

1. Find the volume of the pool

The formula for the volume of a cylinder is  V = πr²h .

D = 12 m; h = 10 m

r = D/2 = (12 m/2) = 6.0 m

V = πr²h = π × (6.0 m)² × 10 m = π × 36 m²× 10 m = 360π m³ = 1100 m³

= 1.1. × 10⁶ L

2. Calculate the moles of OH⁻

n = cV = 1.0 × 10⁻² mol·L⁻¹ × 1.3 × 10⁶ L = 11 000 mol of OH⁻                                                                                                                                                                                                                                                                                                                              

3. Calculate the moles of acetic acid needed for neutralization

HA + OH⁻ ⟶ A⁻ + H₂O

The molar ratio of is 1 mol HA:1 mol OH⁻, so you need 11 000 moL of acetic acid.

4. Calculate the actual moles of acetic acid

You have four 5 L jugs of acetic acid pH 2 .

Volume = 20 L

[H⁺] = 10⁻² mol·L⁻¹ = 0.01 mol·L⁻¹

(a) Set up an ICE table

                      HA + H₂O ⇌ A⁻  + H₃O⁺

I/mol·L⁻¹:          c                   0         0

I/mol·L⁻¹:     - 0.01             +0.01    +0.01

I/mol·L⁻¹:    c - 0.01             0.01     0.01

$K_{\text{a}} = \dfrac{\text{[A]}^{-}\text{[H _{3} O ^{+} ]}}{\text{[HA]}} = 1.76 \times 10^{-5}$

(b) Calculate the concentration of acetic acid

$\begin{array}{rcl}\dfrac{\text{[A]}^{-}\text{[H _{3} O ^{+} ]}}{\text{[HA]}}& = & 1.76 \times 10^{-5}\\\\\dfrac{0.01\times 0.01}{c}& = & 1.76 \times 10^{-5}\\\\1 \times 10^{-4} & = & 1.76 \times 10^{-5}c\\c & = & \dfrac{1 \times 10^{-4}}{1.76 \times 10^{-5}}\\\\ & = & \text{6 mol/L}\\\end{array}$

The concentration of the acetic acid is 6 mol·L⁻¹

(c) Calculate the moles of acetic acid

$n = \text{20 L} \times \dfrac{\text{6 mol}}{\text{1 L}} = \textbf{100 mol}$

You have 100 mol of acetic acid.

The CSI is wrong.

You don't have enough acetic acid to neutralize the pool.