At first sight you might say that, since NaOH is 100% dissociated, then #sf([OH^-]=10^(-7)color(white)(x)"mol/l")#.
Since #sf(pH+pOH=14)# this would mean that such a solution would be #sf(pH=7)# i. e neutral.
We would predict that such a very dilute solution of alkali would have a pH very slightly above 7.
At such a low concentration as this, we must also take into account the ions present from the dissociation of water:
#sf(H_2OrightleftharpoonsH^(+)+OH^(-))#
For which #sf(K_w=[H^+][OH^-]=10^(-14)" ""at"" "25^@C)#
and #sf([H^+]=10^(-7)M)# and #sf([OH^-]=10^(-7)M)#.
To get the total #sf(OH^-)# concentration you might think that this would therefore be #sf(10^(-7)+10^(-7)=2xx10^(-7)color(white)(x)"mol/l")#.
This is not the case. This represents the initial conditions of an equilibrium system which has been disturbed.
The reaction quotient is now #sf(10^(-7)xx2xx10^(-7)=2xx10^(-14))# which exceeds #sf(K_w)#.
Le Chatelier's Principle tells us that the system will relax to restore the postion of equilibrium and the value of #sf(K_w)#.
It will do this by shifting to the left, for which we must set up an ICE table based on #sf("mol/l")#:
#sf(color(white)(xxxx)H_2Ocolor(white)(xxxxx)rightleftharpoonscolor(white)(xxxxx)H^(+)color(white)(xxxxxx)+color(white)(xxxxx)OH^-)#
#sf(Icolor(white)(xxxxxxxxxxxxxxxxxx)10^(-7)color(white)(xxxxxxxxx)(10^(-7)+10^(-7)))#
#sf(Ccolor(white)(xxxxxxxxxxxxxxxxx)-xcolor(white)(xxxxxxxxxxxxxx)-x)#
#sf(Ecolor(white)(xxxxxxxxxxxxxxx)(10^(-7)-x)color(white)(xxxxxxxx)(2xx10^(-7)-x))#
#:.##sf((10^(-7)-x)(2xx10^(-7)-x)=10^(-14))#
This becomes:
#sf(x^(2)-(3xx10^(-7))x+10^(-14)=0)#
If we apply the quadratic formula we get the 2 roots:
#sf(x=2.615xx10^(-7)color(white)(x)"mol/l")#
or
#sf(x=0.385xx10^(-7)color(white)(x)"mol/l")#
I will discard the 1st because this will give a -ve concentration so:
#sf([OH^-]=2xx10^(-7)-0.385xx10^(-7)=1.615xx10^(-7)color(white)(x)"mol/l")#
To get the pH you can say:
#sf(pOH=-log[OH^(-)]=-log[1.615xx10^(-7)]=6.791)#
#sf(pH=14-pOH=14-6.791=7.21)#
Which is the type of result predicted.
