Assuming that this is done under standard conditions we can use the following relationship:
#pH+pOH=14# (at 25 degrees Celcius)
This is crucial since it connects #pH# and #pOH# of a given solution- lets say that the# pH# is 1, then the #pOH# will be 13.
Also remember that:
#pH=-log[H_3O^+]#
#pOH=-log[OH^-]#
In this problem, We can assume that #HCl# is completely ionized in water because #HCl# is a strong acid so the #H_3O^+# concentration will also be #0.015 mol dm^-3#
Then using the equation to find the pH of the given solution:
#pH=-log[H_3O^+]#
#pH=-log[0.015]#
#pH approx 1.82#
#1.82 + pOH=14#
#pOH approx 12.18#
Now we must reverse-engineer the process.
If:
#pOH=-log[OH^-]#
Then:
#10^(-pOH)=[OH^-]#
Hence:
#10^-(12.18)=[OH^-]#
#[OH^-]=10^-(12.18)approx 6.61 times 10^-13 mol dm^-3#
