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Answer 1 · 2017-08-08T15:38:48

$pH=11.2$

We use the relationship $"concentration"$ $=$ $"moles of solute"/"volume of solution"$

And thus $"moles of solute"="concentration"xx"volume"$

And thus ...................

$200*xx10^-3*Lxx1xx10^-3*mol*L^-1=2.00xx10^-4*mol$ with respect to $NaOH$ .

Likewise.....................

$300*xx10^-3*Lxx2xx10^-3*mol*L^-1=6.00xx10^-4*mol$ with respect to $NaOH$ .

And so $[NaOH]=((6.00+2.00)xx10^-4*mol)/((300+200)xx10^-3*L)$

$=1.60xx10^-3*mol*L$ with respect to $NaOH$ ; and of course $[Na^+]=[HO^-]=1.60xx10^-3*mol*L^-1$ .

Note that I have reasonably assumed that the volumes are additive.

Now it is a fact that $pH=-log_10[H_3O^+]$ ......

but given $K_w=10^-14=[H_3O^+][HO^-]$ , we know that......

$pK_w=14=pH+pOH$

From the above..... $pOH=-log_10(1.60xx10^-3)=2.80$

And $pH=14-pOH=14-2.80=11.2$ .........

If you have followed my reasoning, you should be able to tackle most problems that request values of $pH$ and $pOH$ ..........