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

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Answer 1 · 2017-09-13T13:07:55

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

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?

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