How many mL of NaOH are needed to neutralize the acidity in a titrated solution containing .1234 g of #AlCl_3#? Chemistry Reactions in Solution Titration Calculations 1 Answer Andrea B. Jan 30, 2017 Answer: 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# Related questions How do you do acid base titration calculations? How do you use titration calculations to find pH? What is a redox titration and what is it used for? Why is titration used when standardizing a solution? Is titration suitable for sodium nitrate? How can I do redox titration calculations? How can I calculate the titration of a weak acid and a strong base? How can I make back titration calculations? How does titration affect molarity? How does the endpoint of a titration differ from the equivalence point? See all questions in Titration Calculations Impact of this question 570 views around the world You can reuse this answer Creative Commons License