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Acid Base Molarity

Acid-Base Neutralization Reaction

10 Liters 0.03 M NaOH is to be Neutralized with 0.01 M HCl.

How many Liters of the HCl solution will be needed?

Step 1: Write the reaction and balance it

{\it NaOH}+{\it HCl}  -> H_{{2}}O+Na^+ +Cl^-

Step 2: Determine Stoichiometry 

1 NaOH : 1 HCl

Step 3: Determine # of moles of NaOH

10L 0.03M NaOH

M= \frac {mol}{Liters}   moles= M \cdot Liters

moles= 0.03 \cdot 10 = 0.3 moles NaOH

Step 4: Do a conversion with Stoichiomety to get moles HCl needed.

0.3 mols NaOH \cdot {\frac {1 mol \hspace HCl}{1 mol \hspace NaOH}} \cdot 0.3 mol HCl

Step 5: Determine Liters of HCl from the number of moles of HCl.

M= \frac {mol}{L} Liters= \frac {mol}{M} = \frac {0.3\hspace mol \hspace HCl}{0.01 \hspace MHCl} = 30 Liters