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

1 Answer
Mar 2, 2017

Answer:

#[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#.