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

1 Answer
Andrea B.
Jan 30, 2017

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

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