What is the #"normality"# of a solution prepared by dissolving a #5g# mass of sodium hydroxide in a #1L# volume of water?

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Answer 1 · 2017-08-19T14:42:33

#"Normality"=0.125*mol*L^-1#.

Here #"molarity"# #-=# #"normality"#. Why? Because hydroxide anion does not speciate to any degree in aqueous solution.

And #"molarity"# #-=# #"moles of solute"/"volume of solution"#

#=((5*g)/(40.0*g*mol^-1))/(1*L)=(0.125*mol)/(1*L)=0.125*mol*L^-1# with respect to #NaOH(aq)#.

What is #pH# of this solution? We know that #pH+pOH=14# under standard conditions.

What is the #"normality"# of a solution prepared by dissolving a #5*g# mass of sodium hydroxide in a #1*L# volume of water?