Question

How many mL of NaOH are needed to neutralize the acidity in a titrated solution containing .1234 g of `AlCl_3`?

Answer 1

you must know the molarity of NaOH solution!

Explanation:

for `AlCl_3`
`N° mol = G/(MM) =(0,1234g)/ (133,35g/(mol)) = 9,25 10^(-4) mol = 0,925 mmol`
for NaOH
n= M V
n° mol = Molarity x Volume
`V=( N°mol)/ (Molarity)= (0,925 mol)/(x M) = y mL`