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Problems · Problem 3.2

Q.The dissociation constant of NH4OH\mathrm{NH_4OH} is 1.8×10−51.8 \times 10^{-5}. Calculate its degree of dissociation in 0.01 M solution.

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✓ Free question

α=Kb/c=1.8×10−5/0.01=0.04242\alpha = \sqrt{K_b/c} = \sqrt{1.8\times10^{-5}/0.01} = 0.04242 (dimensionless).

Step 1. For the weak base NH4OH\mathrm{NH_4OH}, Ostwald's dilution law in its small-α\alpha form (Eq. 3.11) gives α=Kb/c\alpha = \sqrt{K_b/c}.

Step 2. Substitute Kb=1.8×10−5K_b = 1.8\times10^{-5} and c=0.01=1×10−2c = 0.01 = 1\times10^{-2} M: α=1.8×10−51×10−2=1.8×10−3\alpha = \sqrt{\dfrac{1.8\times10^{-5}}{1\times10^{-2}}} = \sqrt{1.8\times10^{-3}}.

Step 3. Rewrite for a clean even exponent under the root: 1.8×10−3=18×10−41.8\times10^{-3} = 18\times10^{-4}, so α=18×10−2=4.242×10−2\alpha = \sqrt{18}\times10^{-2} = 4.242\times10^{-2}.

Step 4. So α=0.04242\alpha = 0.04242 -- about 4.24% of the NH4OH is dissociated at this concentration.

✓Final answer

α=18×10−4=4.242×10−2=0.04242\alpha = \sqrt{18\times10^{-4}} = 4.242\times10^{-2} = 0.04242 (dimensionless) -- digit-for-digit the textbook's printed final.

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