The pH of a solution containing `NH_(4)OHandNH_(4)^(+)` ia 9. if `[NH_(4)^(+)]=0.1MandKa" of " NH_(4)^(+) " is "5xx10^(-10)` then what is `[NH_(4)OH]`?

Options

A. 0.05 M
B. 20 M
C. 0.5 M
D. 2 M

Answer

To solve the problem step by step, we need to find the concentration of ammonium hydroxide \([NH_4OH]\) in a solution where the pH is given as 9, the concentration of \([NH_4^+]\) is 0.1 M, and the \(K_a\) of \(NH_4^+\) is \(5 \times 10^{-10}\).

### Step 1: Calculate \(K_b\) for \(NH_4OH\)
We know that:
\[ K_w = K_a \times K_b \]
Where \(K_w\) (the ion product of water) at 25°C is \(1 \times 10^{-14}\).

Given:
\[ K_a = 5 \times 10^{-10} \]

We can rearrange the equation to find \(K_b\):
\[ K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{5 \times 10^{-10}} = 2 \times 10^{-5} \]

### Step 2: Calculate \(pOH\) from \(pH\)
We know that:
\[ pH + pOH = 14 \]
Given \(pH = 9\), we can find \(pOH\):
\[ pOH = 14 - pH = 14 - 9 = 5 \]

### Step 3: Calculate the concentration of \(OH^-\) ions
Using the relationship between \(pOH\) and the concentration of hydroxide ions:
\[ pOH = -\log[OH^-] \]
We can find \([OH^-]\):
\[ [OH^-] = 10^{-pOH} = 10^{-5} \, M \]

### Step 4: Set up the equilibrium expression for \(K_b\)
The dissociation of \(NH_4OH\) can be represented as:
\[ NH_4OH \rightleftharpoons NH_4^+ + OH^- \]
The equilibrium expression for \(K_b\) is:
\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]} \]
Substituting the known values:
\[ K_b = \frac{(0.1)(10^{-5})}{[NH_4OH]} \]
Substituting \(K_b = 2 \times 10^{-5}\):
\[ 2 \times 10^{-5} = \frac{(0.1)(10^{-5})}{[NH_4OH]} \]

### Step 5: Solve for \([NH_4OH]\)
Rearranging the equation gives:
\[ [NH_4OH] = \frac{(0.1)(10^{-5})}{2 \times 10^{-5}} = \frac{0.1}{2} = 0.05 \, M \]

### Final Answer
Thus, the concentration of ammonium hydroxide \([NH_4OH]\) is:
\[ [NH_4OH] = 0.05 \, M \]

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