Question

1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

Answer 1

1. Answer:

1.0 × 10–9 M OH–

Explanation:

pH = -Log[H+]

pOH = -Log[OH-]

But;

pH + pOH = 14

Therefore;

[H+] + [OH-] = 1.0 × 10^-14 M

Therefore;

[OH-] = 1.0 × 10^-14 M - (1.0 × 10^–5 M)

         = 1.0 × 10^-9 M OH–

2. Answer;

pH = 7.28

Explanation;

pH = -Log[H3O+]

Given;

[H3O+] = 5.2 × 10^–8 M

Therefore;

pH = - log [5.2 × 10^–8 M]

     = 7.28

The pH is 7.28

Answer 2

Answer :

(1) The $OH^-$ concentration is, $1.0\times 10^{-9}M$

(2) The pH of the solution is, 7.28

Explanation:

Solution for part (1) :

We are given :

Concentration of $H_3O^+$ = $1.0\times 10^{-5}M$

First we have to calculate the pH of the solution.

$pH=-\log [H_3O^+]$

$pH=-\log (1.0\times 10^{-5})$

$pH=5$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-5=9$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$9=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-9}M$

Therefore, the $OH^-$ concentration is, $1.0\times 10^{-9}M$

Solution for part (2) :

We are given :

Concentration of $H_3O^+$ = $5.2\times 10^{-8}M$

Now we have to calculate the pH of the solution.

$pH=-\log [H_3O^+]$

$pH=-\log (5.2\times 10^{-8})$

$pH=7.28$

Therefore, the pH of the solution is, 7.28