A solution contains [OH^-] = 4.0 times 10^-5[OH]=4.0×105 MM, what is the concentration of [H_3O^+][H3O+]?

3 Answers
May 25, 2017

[H_3O^+]=2.50xx10^-10*mol*L^-1[H3O+]=2.50×1010molL1

Explanation:

In aqueous solution, it is a fact that the following equilibrium operates under standard conditions.........

2H_2O(l) rightleftharpoons H_3O^(+) +HO^(-)2H2O(l)H3O++HO

Under standard conditions............

K_"rxn"=K_w=[H_3O^+][HO^-]=10^-14Krxn=Kw=[H3O+][HO]=1014. And typically we would use logarithms to reduce this expression to.......

pH+pOH=14pH+pOH=14, where the pHpH function means -log_10[H_3O^+]log10[H3O+] etc.

So pOH=-log_10[4.0xx10^-5]=4.40pOH=log10[4.0×105]=4.40, and pH=9.60pH=9.60.

And thus [H_3O^+]=10^(-9.60)=2.50xx10^-10*mol*L^-1.[H3O+]=109.60=2.50×1010molL1.

May 25, 2017

[H_3O^+]=2.5*10^(-10) M[H3O+]=2.51010M

Explanation:

Given that no hydrolysis happened in the solution (inferred from question) Calculate [H_3O^+][H3O+] from k_wkw directly.
[H_3O^+][H3O+]
=k_w/([OH^-])=kw[OH]
=10^-14/(4.0*10^-5)=10144.0105
=2.5*10^-10M=2.51010M

May 25, 2017

2.5 xx 10 ^-102.5×1010 "M"M

Explanation:

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