Question

[50 Points] In the experiment below, Beaker 1 contains 12.0 M HCl. Each subsequent beaker is diluted by 50%. What is the [OH-] in Beaker 6?

Answer 1

2.67x10⁻¹⁴M = [OH-]

Step-by-step explanation:

To solve this question we need first to find the HCl concentration of the beaker 6. Then, using:

Kw = [OH-][H+]

Where Kw = 1x10⁻¹⁴

[OH-] is our incognite

[H+] is = [HCl]

Beaker 2 = 12.0M / 2 = 6.0M

Beaker 3 = 6.0M / 2 = 3.0M

Beaker 4 = 3.0M / 2 = 1.5M

Beaker 5 = 1.5M / 2 = 0.75M

Beaker 6 = 0.75M / 2 = 0.375M

HCl = 0.375M = [H+]

1x10⁻¹⁴ / [H+] = [OH-]

1x10⁻¹⁴ / 0.375M = 2.67x10⁻¹⁴M

2.67x10⁻¹⁴M = [OH-]