First we find the pH of solution:
#sf([OH^-]=10^(-3)color(white)(x)M)#
The ionic product of water gives us:
#sf([H^+][OH^-]=10^(-14))#
#:.##sf([H^+]=10^(-14)/[[OH^-]]=10^(-14)/10^(-3)=10^(-11)color(white)(x)M)#
#sf(pH=-log[H^+]=-log[10^(-11)]=11)#
This means we have to create a solution where the pH = 10.
Adding #sf(H^+)# ions will have 2 effects:
-
#sf(OH^-)# ions will be removed due to #sf(H^++OH^(-)rarrH_2O)#. This will lower their concentration.
-
The total volume of the solution will increase. This will also reduce #sf([OH^-])#.
We need to find the volume of acid V to add.
Since the target pH = 10 this means that #sf([H^+]=10^(-10)color(white)(x)M)#
#:.##sf([OH^-]=10^(-14)/10^(-10)=10^(-4)color(white)(x)M)#
The total moles of #sf(OH^-)# in the original solution is given by:
#sf(n_(OH^-)"initial"=cxxv=10^(-3)xx1=10^(-3)=0.001)# mol
The number of moles of #sf(H^+)# added is given by:
#sf(n_(H^+)=cxxV=0.001xxV)# mol
Since they react in the ratio 1:1 the number of moles of #sf(OH^-)# consumed = #sf(0.001xxV)# mol.
So the number of moles of #sf(OH^-)# remaining is given by
#sf(n_(OH^-)=0.001-0.001V)#
In litres the total new volume = #sf((1+V))#
This means the required concentration of #sf(10^(-4)color(white)(x)M)# for #sf(OH^-)# can be written:
#sf([OH^-]=n/v=((0.001-0.001V))/((1+V))=10^(-4)color(white)(x)M)#
#:.##sf((0.001-0.001V)=10^(-4)(1+V))#
#sf(0.001(1-V)=10^(-4)(1+V))#
#sf(((1-V))/((1+V))=10^(-4)/(10^(-3))=0.1)#
#sf((1-V)=0.1(1+V))#
#sf(1-V=0.1+0.1V)#
#sf(1.1V=0.9)#
#sf(V=0.9/1.1=0.818color(white)(x)L)#
#sf(V=818color(white)(x)ml)#
Check by iteration:
#sf([OH^-]=(0.001-0.001xx0.818)/(1+0.818))#
#sf([OH^-]=(cancel(1.818)xx10^(-4))/(cancel(1.818))=10^(-4))#
#:.##sf([H^+]=10^(-14)/([OH^-])=10^(-14)/(10^(-4))=10^(-10)color(white)(x)M)#
#sf(pH=-log[H^+]=-log[10^-10]=10)#
So that's all good.
