Question #d77d9

I will you assume you mean 20 ml and not 20 L. That would be a ridiculous volume for a titration.

Let the general equation be:

`sf(HX+YOHrarrXY+H_(2)O)`

This tells us that 1 mole of HX produces 1 mole of XY.

The number of moles of HX is given by:

`sf(n_(HX)=cxxv=0.2xx20/1000=0.004)`

`:.``sf(n_(XY)=0.004)`

This is the number of moles of XY at the equivalence point.

XY is the salt of a weak acid and a strong base. The solution formed will not be neutral because the `sf(Y^(-))` ion is basic and undergoes hydrolysis:

`sf(Y^(-)+H_2OrightleftharpoonsYH+OH^(-))`

By setting up an ICE table you can obtain a useful expression to get the pOH and hence the pH:

`sf(pOH=1/2[pK_(b)-logb])`

Where b is the concentration of the base i.e `sf([Y^(-)])`

Note that the total volume after the titration = 20 ml + 20 ml = 40 ml

`:.``sf([Y^(-)]=n/v=(0.004)/(40/1000)=0.1color(white)(x)"mol/l")`

To get `sf(pK_b)` we use:

`sf(pK_a+pK_b=pK_w=14)` at `sf(25^@C)`

`:.``sf(pK_(b)=14-pK_(a)=14-5=9)`

`:.``sf(pOH=1/2[9-log(0.1)]=1/2[10]=5)`

We know that:

`sf(pH+pOH=pK_w=14)`

`:.``sf(pH=14-pOH=14-5=9)`

We would expect a result like this as the solution should be alkaline.