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#