Question #31b0a

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Answer 1 · 2017-03-22T05:59:18

Well, this is a basic solution; because it contains MORE of the acidium ion than the hydroxide ion.........

We interrogate the equilibrium:

#2H_2O rightleftharpoons H_3O^+ + HO^-#

For which #K_w=10^(-14)=[H_3O^+][HO^-]#

Taking #log_10# of both sides:

#log_(10)10^-14=log_10[H_3O^+]+log_10[HO^-]#

And on rearrangement, #14=-log_10[H_3O^+]-log_10[HO^-]#,

i.e. #14=pH+pOH#.

#"Concentration of sodium hydroxide"#

#=("Moles of sodium hydroxide")/("Volume of solution")#

#=# #((0.800*g)/(39.998*g*mol^-1))/(2.00*L)=1.00xx10^-2*mol*L^-1#

Now, clearly, we can take the #pH# of this solution, given that we know that #pH+pOH=14#, and #pOH=-log_10[HO^-]#
#=-log_10(1.00xx10^-2)=2.0#

Thus #pH=14-2=12#......clearly an alkaline solution,