Question

Acid & Base Calculations Calculate the hydronium ion concentration for each. Tell whether it is an acid or a base. 1. pH = 5.54 ____________ ________ 2. pOH = 9.7 ____________ ________ 3. pH = 7.0 ____________ ________ 4. pH = 12.9 ____________ ________ 5. pOH = 1.2 ____________ ________ Calculate the pOH for each. Tell whether it is an acid or a base. 6. [H3O+] = 1 x 10-5 M ________ ________

Answer 1

Explanation:

pH is the negative logarithm of hydronium ion concentration present in a solution.

• If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9
• If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
• The solution having pH equal to 7 is termed as neutral solution.

To calculate the pH of the solution, we use equation:

$pH=-\log[H_3O^+]$     ......(1)

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14                ........(2)

• For 1:

We are given:

pH = 5.54

Putting values in equation 1, we get:

$5.54=-\log[H_3O^+]$

$[H_3O^+]=2.88\times 10^{-6}M$

Now, putting values in equation 2, we get:

14 = 5.54 + pOH

pOH = 8.46

The solution is acidic in nature.

• For 2:

We are given:

pOH = 9.7

Putting values in equation 2, we get:

14 = 9.7 + pH

pH = 4.3

Now, putting values in equation 1, we get:

$4.3=-\log[H_3O^+]$

$[H_3O^+]=5.012\times 10^{-5}M$

The solution is acidic in nature.

• For 3:

We are given:

pH = 7.0

Putting values in equation 1, we get:

$7.0=-\log[H_3O^+]$

$[H_3O^+]=1.00\times 10^{-7}M$

Now, putting values in equation 2, we get:

14 = 7.0 + pOH

pOH = 7.0

The solution is neither acidic nor basic in nature.

• For 4:

We are given:

pH = 12.9

Putting values in equation 1, we get:

$12.9=-\log[H_3O^+]$

$[H_3O^+]=1.26\times 10^{-13}M$

Now, putting values in equation 2, we get:

14 = 12.9 + pOH

pOH = 1.1

The solution is basic in nature.

• For 5:

We are given:

pOH = 1.2

Putting values in equation 2, we get:

14 = 1.2 + pH

pH = 12.8

Now, putting values in equation 1, we get:

$12.8=-\log[H_3O^+]$

$[H_3O^+]=1.58\times 10^{-13}M$

The solution is basic in nature.

• For 6:

We are given:

$[H_3O^+]=1\times 10^{-5}M$

Putting values in equation 1, we get:

$pH=-\log(1\times 10^{-5})$

$pH=5$

Now, putting values in equation 2, we get:

14 = 5 + pOH

pOH = 9

The solution is acidic in nature.