This is a very low concentration so we must take into account the dissociation of water rather than use #sf(10^(-7))# as the #sf(H^(+))# concentration, which would give a pH of 7.
Water dissociates:
#sf(H_2OrightleftharpoonsH^(+)+OH^(-))#
#sf(K_w=[H^(+)][OH^(-)]=10^(-14))# at #sf(25^@C)#
If we assume a tiny amount of HCl is added then we have disturbed a system at equilibrium. The system will react as to oppose this change and shift to the left to give a new position of equilibrium.
The initial concentration of #sf(H^+)# will be #sf(10^(-7)M)# from the water + #sf(10^(-7)M)# from the HCl = #sf(2xx10^(-7)M)#
We can use an ICE table based on mol/l to show this:
#sf(color(white)(xxxx)H_2Ocolor(white)(xxxxx)rightleftharpoonscolor(white)(xxxxxx)H^(+)color(white)(xxxxxx)+color(white)(xxxxxxx)OH^(-))#
#sf(Icolor(white)(xxxxxxxxxxxxxxxxxx)2xx10^(-7)color(white)(xxxxxxxxxxxxx)10^(-7))#
#sf(Ccolor(white)(xxxxxxxxxxxxxxxxxx)-xcolor(white)(xxxxxxxxxxxxxxx)-x)#
#sf(Ecolor(white)(xxxxxxxxxxxxxxx)(2xx10^(-7)-x)color(white)(xxxxxxxx)(10^(-7)-x))#
#:.##sf((2xx10^(-7)-x)(10^(-7)-x)=10^(-14))#
This gives:
#sf(x^(2)-(3xx10^(-7))x+10^(-14)=0)#
If we apply the quadratic formula, ignoring the absurd root we get:
#sf(x=0.385xx10^(-7)color(white)(x)"mol/l")#
#:.##sf([H^+]=2xx10^(-7)-0.385xx10^(-7)=1.615xx10^(-7)color(white)(x)"mol/l")#
#sf(pH=-log[H^+]=-log[1.615xx10^(-7)]=6.79)#
As we would expect, the solution is very slightly acidic.
#sf(pH+pOH=14)#
#:.##sf(pOH=14-6.79=7.208)#
