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Step-by-Step Solution
Step 1: Calculate moles of NH4Cl
The given concentration of NH4Cl is 0.0504 M and the volume used is 5.0 mL (which is 5.0 × 10−3 L).
Moles of NH4Cl = (Concentration) × (Volume in L)
= 0.0504 × (5.0 × 10−3)
= 2.52 × 10−4 moles.
Step 2: Calculate moles of NH3
The given concentration of NH3 is 0.0210 M and the volume used is 2.0 mL (which is 2.0 × 10−3 L).
Moles of NH3 = (Concentration) × (Volume in L)
= 0.0210 × (2.0 × 10−3)
= 4.20 × 10−5 moles.
Step 3: Calculate total volume of the mixture
The total volume of the mixture is the sum of the two volumes:
Total volume = 5.0 mL + 2.0 mL = 7.0 mL = 7.0 × 10−3 L.
Step 4: Calculate the final concentrations of NH4+ and NH3
Since NH4Cl dissociates completely, the moles of NH4+ is the same as the moles of NH4Cl.
[NH_4^+] = \frac{\text{moles of } NH_4Cl}{\text{total volume}}
= \frac{2.52 \times 10^{-4}}{7.0 \times 10^{-3}}
\approx 0.036 \text{ M}
Similarly, NH3 is not ionized completely, but its initial concentration in the mixture is:
[NH_3] = \frac{\text{moles of } NH_3}{\text{total volume}}
= \frac{4.20 \times 10^{-5}}{7.0 \times 10^{-3}}
\approx 0.006 \text{ M}
Step 5: Use the expression for Kb
For the base dissociation equilibrium:
K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}
Step 6: Solve for [OH−]
Rearranging,
[OH^-] = \frac{K_b \times [NH_3]}{[NH_4^+]}
Substitute the values:
K_b = 1.8 \times 10^{-5}, \quad [NH_3] \approx 0.006, \quad [NH_4^+] \approx 0.036
[OH^-] = \frac{1.8 \times 10^{-5} \times 0.006}{0.036}
Simplify step by step:
1) 1.8 \times 10^{-5} \times 0.006 = 1.08 \times 10^{-7}
2) 1.08 \times 10^{-7} \div 0.036 = 3.0 \times 10^{-6}
So, [OH^-] \approx 3.0 \times 10^{-6}\,\text{M}.
Step 7: Identify the value of x
The problem states that the hydroxide concentration is x \times 10^{-6} . From our calculation,
[OH^-] = 3.0 \times 10^{-6}\,\text{M}.
So, x = 3.
