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

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Answer 1 · 2017-03-02T20:13:30

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

We assume an aqueous solution under standard conditions.

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