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""normality" of a solution prepared by dissolving a 5*g5*g mass of sodium hydroxide in a 1*L1*L volume of water?