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The acid ionziation (hydrolysis) constant of $$Zn^{2+}$$ is $$1.0\times 10^{-9}$$
(a) Calculate the $$pH$$ of a $$0.001\ M$$ solution of $$ZnCl_{2}$$
(b) What is the basic dissociation constant of $$Zn(OH)^{+}$$?


Solution

(Acid ionisation) hydrolysis constant of $$Zn^{2+}=1.0\times 10^{-9}$$
Let it be $$x \implies x=10^{-9}$$        also     $$c=0.001$$
We know, $$x=\sqrt{\cfrac{K_w}{K_b .C}}=\sqrt{\cfrac{10^{-14}}{10^{-3}.K_b}}=10^{-9}$$
$$\cfrac { { 10 }^{ -14 } }{ { 10 }^{ -3 }.{ K }_{ b } } ={ 10 }^{ -18 }=\cfrac { 1 }{ { 10 }^{ -7 } } ={ K }_{ b }$$
$$K_b=10^7$$
$$pK_b=-log10^7=-7$$
$$pH=\cfrac 1 2 (14-pK_b-logC)$$
$$=\cfrac 1 2(14+7-log10^{-3})=\cfrac{24}{2}=12$$
$$pH=12$$

Chemistry

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