Suppose 1.2 grams of calcium hydroxide are dissolved in 1.75 L of an aqueous solution. How do you determine the following: hydroxide ion concentration, hydronium ion concentration, calcium ion concentration, pH and poH?

1 Answer
anor277
Jun 1, 2017

"Concentration"="Amount of stuff"/"Volume of solution"Concentration=Amount of stuffVolume of solution

Explanation:

And generally, we report "concentration"concentration with units of mol*L^-1mol⋅L−1.

And so [Ca(OH)_2]=((1.2*g)/(74.09*g*mol^-1))/(1.75*L)=9.26xx10^-3*mol*L^-1[Ca(OH)2]=1.2⋅g74.09⋅g⋅mol−11.75⋅L=9.26×10−3⋅mol⋅L−1.

Clearly, [HO^-]=2xx9.26xx10^-3*mol*L^-1[HO−]=2×9.26×10−3⋅mol⋅L−1. Agreed? Because [Ca(OH)_2][Ca(OH)2] speciates in solution to give Ca^(2+)Ca2+, and 2 equiv of hydroxide ion.

And thus [Ca^(2+)]=9.26xx10^-3[Ca2+]=9.26×10−3, [HO^(-)]=1.85xx10^-2*mol*L^-1[HO−]=1.85×10−2⋅mol⋅L−1.

pOHpOH is simply a measure of hydroxide ion concentration, and in aqueous solution we know that pH+pOH=14pH+pOH=14.

pOH=-log_10[HO^-]=-log_10(1.85xx10^-2)=1.73pOH=−log10[HO−]=−log10(1.85×10−2)=1.73

pH=14-1.73=12.27pH=14−1.73=12.27.

For another example of this sort of problem, see [here.](https://socratic.org/questions/is-a-solution-with-a-poh-of-12-1-acidic-basic-or-neither)

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