Question

What is `pH` of an aqueous solution that is `0.016*mol*L^-1` with respect to `NaOH`?

Answer 1

`pH=12.2`

Explanation:

In aqueous solution, `pH=-log_(10)[H_3O^+]`, and likewise, `pH=-log_(10)[HO^-]`.

Now since we interrogate the equilibrium:

`2H_2Orightleftharpoons H_3O^+ + ""^(-)OH` where `K_"water"=10^(-14)`, then..........

`[H_3O^+][HO^-]=10^(-14)`, and we take `log_10` of both sides......

`log_10[H_3O^+]+ log_10[HO^-]=log_(10)10^(-14)=-14`

On rearrangement,

`14=-log_10[H_3O^+]-log_10[HO^-]=pH+pOH`.

And this is our defining relationship, `pH+pOH=14` (and this has to be known and used!)

So, we find the `pOH` of a solution that is `0.016*mol*L^-1` with respect to `NaOH`:

`pOH-=-log_(10)(0.016)=1.80`

And finally,

`pH=14-pOH=14-1.80=12.2`

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