Question

Calculate the pH N/1000 NaOH solution assuming complete ionisation???

Answer 1

as normality = molarity ×n factor
then molarity of NaOH=.oo1molar
then pOH=-log[OH-]= - log 10 raised to the power -3
therefore pOH=3
then pH+pOH=14
pH=11

Answer 2

Answer: 11

Explanation:

$NaOH\rightarrow Na^++OH^-$

$Normality= Molarity\times n$

n= acidity of $NaOH$ = number of hydroxyl ions produced in water = 1

Thus $Normality= Molarity$

Thus Molarity is 0.001 M.

Thus $[OH^-]$= 0.001 M

$pOH=-log[OH^-]=-log[10^{-3}]$

$pOH=3$

$pH+pOH=14$

$pH+3=14$

$pH=11$