Question #ed4f1

1 Answer
Aug 8, 2017

Answer:

#pH=11.2#

Explanation:

We use the relationship #"concentration"# #=# #"moles of solute"/"volume of solution"#

And thus #"moles of solute"="concentration"xx"volume"#

And thus ...................

#200*xx10^-3*Lxx1xx10^-3*mol*L^-1=2.00xx10^-4*mol# with respect to #NaOH#.

Likewise.....................

#300*xx10^-3*Lxx2xx10^-3*mol*L^-1=6.00xx10^-4*mol# with respect to #NaOH#.

And so #[NaOH]=((6.00+2.00)xx10^-4*mol)/((300+200)xx10^-3*L)#

#=1.60xx10^-3*mol*L# with respect to #NaOH#; and of course #[Na^+]=[HO^-]=1.60xx10^-3*mol*L^-1#.

Note that I have reasonably assumed that the volumes are additive.

Now it is a fact that #pH=-log_10[H_3O^+]#......

but given #K_w=10^-14=[H_3O^+][HO^-]#, we know that......

#pK_w=14=pH+pOH#

From the above.....#pOH=-log_10(1.60xx10^-3)=2.80#

And #pH=14-pOH=14-2.80=11.2#.........

If you have followed my reasoning, you should be able to tackle most problems that request values of #pH# and #pOH#..........