Equal volumes of 0.02 M CaCl_2 and 0.00004 | vedclass

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(B) When equal volumes are mixed,the total volume doubles,so the concentration of each species is halved. $[Ca^{2+}] = \frac{0.02}{2} = 0.01 \ M = 10^

Vedclass · Accepted Answer

No,precipitation will not occur.

Vedclass · Answer

(B) When equal volumes are mixed,the total volume doubles,so the concentration of each species is halved. $[Ca^{2+}] = \frac{0.02}{2} = 0.01 \ M = 10^{-2} \ M$. $[SO_4^{2-}] = \frac{0.00004}{2} = 0.00002 \ M = 2 	imes 10^{-5} \ M$. The ionic product $Q_{sp} = [Ca^{2+}][SO_4^{2-}] = (10^{-2}) 	imes (2 	imes 10^{-5}) = 2 	imes 10^{-7}$. Since $Q_{sp} (2 	imes 10^{-7}) < K_{sp} (2.4 	imes 10^{-5})$,the solution is unsaturated and no precipitation of $CaSO_4$ will occur.

Equal volumes of $0.02 \ M$ $CaCl_2$ and $0.00004 \ M$ $Na_2SO_4$ solutions are mixed. Will a precipitation of $CaSO_4$ occur? $(K_{sp} = 2.4 \times 10^{-5})$

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