Question

A `0.200*g` mass of sodium hydroxide is dissolved in a `0.500*L` volume of water? What are `pOH`, and `pH` of this solution?

Answer 1

`pH=14-pOH`

Explanation:

`[HO^-]="Moles of hydroxide"/"Volume of solution"`

`=((0.2*g)/(40.00*g*mol^-1))/(0.500*L)=0.01*mol*L^-1`.

We (reasonably) assume no volume change on dissolution. Some heat will be evolved. Why?

`pOH=-log_10[HO^-]=-log_10(10^-2)=-(-2)=2`.

But we know for a fact that `pH+pOH=14`

And thus `pH=............?`