Question #52060

2 Answers
May 18, 2017

Answer:

Explanation:

I assume you mean #sf(10^-8color(white)(x)M)# NaOH solution.

The pH cannot be 6 because, as you correctly state, NaOH is a base or alkali. I guess you reasoned:

#sf(K_w=[H^+][OH^-]=10^(-14)color(white)(x)"mol"^2."l"^(-2))# at 298K

#:.##sf([H^+]=10^(-14)/[[OH^-]]=10^(-14)/10^(-8)=10^(-6)color(white)(x)"mol/l")#

#sf(pH=-log[H^+]=-log(10^-6)=6)#

Clearly, this cannot be right as such a solution would be weakly acidic.

#sf(10^-8)#M is a fantastically dilute solution so you would expect the pH to fall towards 7 as the solution is diluted. It cannot fall below 7.

At these low concentrations you need to take into account the #sf(H^+)# formed from the auto - ionisation of water. This is done in the answer which I have given the link to.

May 18, 2017

Answer:

The #pH# is indeed #6.0#.

Explanation:

You can find the #pH# from a given #[OH^-]# using an equation like this:

#pH = 14.0-(-log([OH^-]))#

#pH = 14.0-(-log(10^-8 M))#

#pH = 14.0-(8.0)= **6.0** #

#NaOH# is a base, but the solution is basic only if the #[OH^-]# is greater than #1 xx 10^-7 M#, i.e. greater than #[H^+]#.