A solution contains #[OH^-] = 4.0 times 10^-5# #M#, what is the concentration of #[H_3O^+]#?

3 Answers
May 25, 2017

Answer:

#[H_3O^+]=2.50xx10^-10*mol*L^-1#

Explanation:

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

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

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

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

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

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

And thus #[H_3O^+]=10^(-9.60)=2.50xx10^-10*mol*L^-1.#

May 25, 2017

Answer:

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

Explanation:

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

May 25, 2017

Answer:

#2.5 xx 10 ^-10# #"M"#

Explanation:

My own work (Mert METÄ°N)