To determine the molarity of NH₄Br given the pH of the solution and the molarity of NH₃, let's break down the problem step-by-step.
### Step 1: Determine the pOH from the given pH
The pH of the solution is provided as 8.3. The relationship between pH and pOH is given by:
[tex]\[ \text{pOH} = 14 - \text{pH} \][/tex]
Substituting the given pH value:
[tex]\[ \text{pOH} = 14 - 8.3 = 5.7 \][/tex]
### Step 2: Calculate the hydroxide ion concentration [OH⁻] from the pOH
The concentration of hydroxide ions [OH⁻] can be found using the equation:
[tex]\[ [\text{OH}^-] = 10^{-\text{pOH}} \][/tex]
Substituting the calculated pOH value:
[tex]\[ [\text{OH}^-] = 10^{-5.7} \][/tex]
This calculation yields:
[tex]\[ [\text{OH}^-] \approx 1.9952 \times 10^{-6} \, \text{M} \][/tex]
### Step 3: Use the ammonia equilibrium constant (Kb) to find the concentration of NH₄⁺
The given equilibrium constant [tex]\( K_b \)[/tex] for NH₃ (ammonia) is:
[tex]\[ K_b = 1.8 \times 10^{-5} \][/tex]
For the equilibrium reaction:
[tex]\[ \text{NH}_3 + \text{H}_2\text{O} \leftrightarrow \text{NH}_4^+ + \text{OH}^- \][/tex]
The equilibrium expression is:
[tex]\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \][/tex]
Rearranging to solve for the concentration of NH₄⁺, we get:
[tex]\[ [\text{NH}_4^+] = \frac{K_b \times [\text{NH}_3]}{[\text{OH}^-]} \][/tex]
Substituting the known values:
[tex]\[ [\text{NH}_4^+] = \frac{1.8 \times 10^{-5} \times 0.016}{1.9952 \times 10^{-6}} \][/tex]
After solving the equation, we find:
[tex]\[ [\text{NH}_4^+] \approx 0.00177 \, \text{M} \][/tex]
### Step 4: Express the final answer to two significant figures
Since the solution requires the molarity of NH₄Br, we note that in solution, NH₄Br dissociates completely to provide NH₄⁺ ions, so:
[tex]\[ [\text{NH}_4\text{Br}] = [\text{NH}_4^+] = 0.0018 \, \text{M} \][/tex]
Therefore, the molarity of NH₄Br is:
[tex]\[ \boxed{0.0018 \, \text{M}} \][/tex]
