What is the #H_3O^+# ion concentration of a solution whose #OH^- # ion concentration is # 1 xx 10^-3 M#?

1 Answer
Jun 19, 2017

Answer:

We know that #pH+pOH=14#..........so #pH=11#, and #[H_3O^+]=10^-11*mol*L^-1#.

Explanation:

We use the following equilibrium..........

#H_2OrightleftharpoonsH^+ +HO^(-)#, and further that we know the equilibrium constant for this expression is......

#[H^+]xx[HO^-]=10^-14# under standard conditions.

Alternatively, #[H_3O^+][HO^-]=10^-14#. We can take #log_10# of both sides and get.........

#log_10[H_3O^+]+log_10[HO^-]=log_10(10^-14)#,

or......#underbrace(-log_10(10^-14))_(pK_w)=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)#,

or......#-log_10(10^-14)=pH+pOH=14#

But the expression on the left hand side is simply #14#, and this goes back to definition of the log function, the #log# of something is the exponent to which you raise the #"base"# (normally #10# or #e#) to get the bracketed term.

or......#-log_10(10^-14)=-(-14)=14#

And so finally #pH+pOH=14#, you must know this for A-level....

Here #[HO^-]=10^-3*mol*L^-1#, so #pOH=-log_10(10^-3)=-(-3)=3#

#pH=11#, and #pOH=3#, i.e. taking antilogs.......

#[H_3O^+]=10^-11*mol*L^-1.#