The pH of aqueous solution of ammonia is 11.5. Find molarity of solution. `K_b (NH_4OH)= 1.8 xx 10^(-5)`

Answer

To find the molarity of the ammonia solution given that the pH is 11.5 and the base dissociation constant \( K_b \) for \( NH_4OH \) is \( 1.8 \times 10^{-5} \), we can follow these steps:

### Step 1: Calculate the concentration of \( H^+ \) ions
The pH of the solution is given as 11.5. We can find the concentration of hydrogen ions (\( H^+ \)) using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-11.5} \]
Calculating this gives:
\[ [H^+] = 3.16 \times 10^{-12} \, \text{M} \]

### Step 2: Calculate the concentration of \( OH^- \) ions
Using the relationship between \( H^+ \) and \( OH^- \) ions, we can find the concentration of hydroxide ions (\( OH^- \)):
\[ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \]
Rearranging gives:
\[ [OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{3.16 \times 10^{-12}} \]
Calculating this gives:
\[ [OH^-] = 3.16 \times 10^{-3} \, \text{M} \]

### Step 3: Set up the equilibrium expression for \( K_b \)
The dissociation of ammonia in water can be represented as:
\[ NH_4OH \rightleftharpoons NH_4^+ + OH^- \]
The expression for the base dissociation constant \( K_b \) is given by:
\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]} \]
Assuming \( x \) is the molarity of the ammonia solution, at equilibrium:
\[ [NH_4^+] = [OH^-] = 3.16 \times 10^{-3} \, \text{M} \]
And since \( NH_4OH \) is a weak base, we can approximate:
\[ [NH_4OH] \approx x - 3.16 \times 10^{-3} \approx x \quad (\text{since } x \gg 3.16 \times 10^{-3}) \]

### Step 4: Substitute values into the \( K_b \) expression
Substituting the known values into the \( K_b \) expression:
\[ 1.8 \times 10^{-5} = \frac{(3.16 \times 10^{-3})(3.16 \times 10^{-3})}{x} \]
This simplifies to:
\[ 1.8 \times 10^{-5} = \frac{(1.0 \times 10^{-5})}{x} \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ x = \frac{1.0 \times 10^{-5}}{1.8 \times 10^{-5}} \approx 0.556 \, \text{M} \]

### Conclusion
The molarity of the ammonia solution is approximately \( 0.556 \, \text{M} \).

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