Question

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

Answer 1

`[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?