A solution of ammonia has a pH of 11.8. What is the concentration of [tex][tex]$OH^{-}$[/tex][/tex] ions in the solution?

Useful formulas include:
[tex]\left[ H_3O^+ \right] = 10^{-pH}[/tex],
[tex]\left[ OH^{-} \right] = 10^{-pOH}[/tex],
[tex]pH + pOH = 14[/tex], and
[tex]\left[ H_3O^+ \right] \left[ OH^{-} \right] = 10^{-14}[/tex].

A. [tex]1.58 \times 10^{-12} \, M[/tex]
B. [tex]6.31 \times 10^{-3} \, M[/tex]
C. 2.20 M
D. 158 M



Answer :

To determine the concentration of [tex]\( OH^- \)[/tex] ions in an ammonia solution with a pH of 11.8, we will follow several steps using the provided formulas.

### Step 1: Calculate pOH
First, we use the relation between pH and pOH:

[tex]\[ \text{pH} + \text{pOH} = 14 \][/tex]

Given that the pH is 11.8, we can calculate the pOH as follows:

[tex]\[ \text{pOH} = 14 - \text{pH} \][/tex]
[tex]\[ \text{pOH} = 14 - 11.8 \][/tex]
[tex]\[ \text{pOH} = 2.2 \][/tex]

### Step 2: Calculate [tex]\([OH^-]\)[/tex] ion concentration
Next, we use the formula that relates pOH to the concentration of hydroxide ions [tex]\([OH^-]\)[/tex]:

[tex]\[ [OH^-] = 10^{-\text{pOH}} \][/tex]

With pOH = 2.2, the concentration of [tex]\( OH^- \)[/tex] is:

[tex]\[ [OH^-] = 10^{-2.2} \][/tex]

The value [tex]\( 10^{-2.2} \)[/tex] can be calculated precisely:

[tex]\[ 10^{-2.2} \approx 0.00631 \][/tex]

### Step 3: Interpret the result
Hence, the concentration of [tex]\( OH^- \)[/tex] ions in the solution is approximately [tex]\( 0.00631 \, M \)[/tex].

So, the correct answer is:

[tex]\[ 6.31 \times 10^{-3} \, M \][/tex]