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.