Question

Complete the following table.

[ H3O+ ]	[ OH− ]	pH	pOH	Acidic or Basic?
a) 6.4 × 10−3	× 10(select)−12−16−10−18−14−8
b) ___ × 10(____)	9.3 × 10−3
c) _____ × 10(__)	× 10(select)−4−10−14−6−8−12		11.68
d) ____ × 10(__)	× 10(select)−3−11−73−5	9.05
e) ____ × 10( __ )	2.3 × 10−8

Answer 1

the formulas that is use ful is

[H3O+] [OH-] = 1.0 x 10^-14

pH = -log[H3O+]

pOH = -log[OH-]

pH + pOH = 14

Part A

[H3O+] = 6.4 × 10⁻³

[OH-] = 1.0 x 10^-14 / 6.4 × 10⁻³ = 1.56 x 10^-12

pH = -log(6.4 × 10⁻³) = 2.19

pOH = 14-2.19 = 11.8

acidic

Part b

[OH-] = 9.3 × 10⁻³

[H3O+] = 1.0 x 10^-14 / 9.3 × 10⁻³ = 1.07 x 10^-12

pH = -log[1.07 x 10^-12] = 11.97

pOH = 14 - 11.97 = 2.03

Basic

Part C

pOH = 11.68

pH = 14-11.68 = 2.32

[H3O+] = 10^-pH = 10^-2.32 = 4.8 x 10^-3

[OH-] = 10^-pOH = 10^-11.68 = 2.1 x 10^-12

acidic

Part D

pH = 9.05

pOH = 14 - 9.05 = 4.95

[H3O+] = 10^-pH = 10^-9.05 = 8.9 x 10^-10

[OH-] = 10^-pOH = 10^-4.95 = 1.12 x 10^-5

basic

Part e

[OH-] = 2.3× 10⁻⁸

[H3O+] = 1.0 x 10^-14 /2.3 × 10⁻⁸ = 4.34 x 10^-7

pH = -log[4.34 x 10^-7] = 6.36

pOH = 14 - 6.36 = 7.63

slightly acidic (or nuetral)