What is the $"pH"$ of $"0.1 N"$ $"Ca"("OH")_2$ ? Thanks.

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Answer 1 · 2017-07-01T15:46:07

$pH=13.0$

We work out (i) the $pOH$ ........

We have $0.1*N$ with respect to $HO^-$ . This is $0.05*mol*L^-1$ with respect to $Ca(OH)_2(aq)$ . Agreed?

Now.................

$pOH=-log_10[HO^-]=-log_10(0.1)=-(-1)=+1$ .

And we can now (ii) address the $pH$ .

Because $pOH+pH=14$ , and thus $pH=14-1.00=13.0$ .

Is this solution acidic or basic?

See here for a similar problem.

Thanks to Truong-Son who pointed out an error on my part......

What is the "pH""pH" of "0.1 N""0.1 N" "Ca"("OH")_2"Ca"("OH")_2? Thanks.