A #0.200*g# mass of sodium hydroxide is dissolved in a #0.500*L# volume of water? What are #pOH#, and #pH# of this solution?

1 Answer
Jan 4, 2017

Answer:

#pH=14-pOH#

Explanation:

#[HO^-]="Moles of hydroxide"/"Volume of solution"#

#=((0.2*g)/(40.00*g*mol^-1))/(0.500*L)=0.01*mol*L^-1#.

We (reasonably) assume no volume change on dissolution. Some heat will be evolved. Why?

#pOH=-log_10[HO^-]=-log_10(10^-2)=-(-2)=2#.

But we know for a fact that #pH+pOH=14#

And thus #pH=............?#