If #pH=1.65#, what is #[H_3O^+]#?

1 Answer
Sep 13, 2017

Answer:

#[H^+]# or #[H_3O^+]=0.0224*mol*L^-1#

Explanation:

For the equilibrium,

#2H_2O rightleftharpoonsH_3O^+ + HO^-#...

alternatively, #H_2O rightleftharpoons H^+ + HO^-#

#K_"eq"=[H_3O^+][HO^-]=10^-14# under standard conditions.....

And we take #log_10# of BOTH SIDES, and gets....

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

#underbrace(-log_10[H_3O^+])_(pH+)underbrace(-log_10[HO^-])_(pOH)=+14#

And so #pH+pOH=14#, under standard conditions, the which we assume.

And thus for #[H]^+#/#[H_3O^+]#, we take #10^(-1.65)*mol*L^-1=0.0224*mol*L^-1#.

What is #[HO^-]# for this solution?