Question

If the H ion concentration is 0.00075 M, what is the OH ion concentration?

Answer 1

`[HO^-]=10^(-10.88)=1.33xx10^-11*mol*L^-1`.

We assume an aqueous solution under standard conditions.

Explanation:

We know that `[HO^-][H_3O^+]=K_w=10^-14`

If we take `log_10` of both sides, then:

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

But by definition, when I say `log_ab=c`, I ask for the power to which I raise the base `a` to get `b`. So since `log_ab=c`, then `a^c=b`.

And likewise, `log_10[10^-14]=-14`, because clearly the exponent of `10^-14=-14`.

And so `log_10[HO^-]+log_10[H_3O^+]=log_10[10^-14]=-14`

OR,

`14=-log_10[HO^-]-log_10[H_3O^+]`,

i.e. `14=pH+pOH`, because that is how we define `pH` etc., i.e. `pH=-log_10[H_3O^+]`.

We are given that `[H_3O^+]=0.00075*mol*L^-1`

And `pH=-log_10(0.00075)=-(-3.13)=3.13`

So `pOH=10.88`.

And `[HO^-]=10^(-10.88)=1.33xx10^-11*mol*L^-1`.