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Find moles of `NH_(4)Cl` required to prevent `Mg(OH)_(2)` from precipitating in a litre of solution which contains 0.02 mole `NH_(3)` and 0.001 mole `Mg^(2+)` ions. Given : `K_(b)(NH_(3))=10^(-5),` `K_(sp)[Mg(OH)_(2)]=10^(-11)`.A. `10^(-4)`B. `2 xx 10^(-3)`C. `0.02`D. `0.1` |
Answer» Correct Answer - B `[OH^(-)]= sqrt((K_(SP))/([Mg^(2+)])) = sqrt((10^(-11))/(10^(-3))) = 10^(-4)` `rArr pOH = pK_(b) + "log" ([NH_(4)Cl])/([NH_(3)])` |
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