What is the [OH^-][OH−] of a 4.0 times 10^-44.0×10−4 MM solution of Ca(OH_2)Ca(OH2)?

1 Answer
anor277
Apr 7, 2017

[HO^-]=8.0xx10^-4*mol*L^-1[HO−]=8.0×10−4⋅mol⋅L−1............

Explanation:

Calcium hydroxide speciates in aqueous solution according to the following reaction:

Ca(OH)_2(s) stackrel(H_2O)rarrCa^(2+) + 2HO^-Ca(OH)2(s)H2O−−→Ca2++2HO−

If the the concentration is 4.0xx10^-4*mol*L^-14.0×10−4⋅mol⋅L−1 with respect to Ca^(2+)Ca2+, stoichiometry, i.e. the composition of the salt, demands the solution is TWICE this concentrated with respect to ""^(-)OH−OH.

And thus [HO^-]=2xx[Ca^(2+)]=2xx4.0xx10^-4*mol*L^-1=??*mol*L^-1.[HO−]=2×[Ca2+]=2×4.0×10−4⋅mol⋅L−1=??⋅mol⋅L−1.

What are pOHpOH and pHpH of this solution?

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