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)
