If $[HO^-]=6.0xx10^-10molL^-1$ , what is $pH$ for this solution?

Question

2017 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-31T14:32:21

$[H^+] ("or "[H_3O^+])=10^(-4.78)*mol*L^-1$ , $pH=4.78$ , $pOH=9.22$

In aqueous solution at $298*K$ , water undergoes the following equilibrium reaction:

$H_2O(l) rightleftharpoons H^(+) + HO^(-)$

Alternatively, and perhaps now more commonly we would write:

$2H_2O(l) rightleftharpoons H_3O^+ + HO^-$

This equilibrium reaction has been carefully and quantitatively measured, and under standard conditions:

$[H_3O^+][HO^-]=10^-14$

If we take negative logarithms to the base 10 of both sides we get:

$pK_w=14=pOH+pH$ , where the $pH$ functions means $-log_10[H_3O^+]$ , etc.

We are given $[HO^-]=6.0xx10^-10$ , so $pOH=-log_10(6.0xx10^-10)=-(-9.22)=9.22$ , and thus $pH=14-9.22=4.78$ .

Is this solution acidic or basic?

The given reaction occurs at $298*K$ . Given that this is a bond-breaking reaction, how do you think $pK_w$ would evolve at $323*K$ or $373*K$ ?

If [HO^-]=6.0xx10^-10*mol*L^-1[HO^-]=6.0xx10^-10*mol*L^-1, what is pHpH for this solution?