Calculate pH and pOH of 1.0*10^-7 HCl solution? (hint;use quadratic equation)

1 Answer
Apr 23, 2018

pH = 6.79

pOH = 7.21

Explanation:

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)#