Calculate the molar concentration

Question

Calculate the molar concentration of the acid if 35.18mL of hydrochloric acid was required to neutralize 45.0 mL of 0.345 M aluminum hydroxide.

Answer 1

Write a correctly balanced equation:

3HCl(aq) + Al(OH)₃(aq) ===> 3H₂O(l) + AlCl₃(aq)

moles Al(OH)₃ present = 45.0 ml x 1 L/1000 ml x 0.345 moles/L = 0.01553 moles

moles HCl required for neutralization = 0.01553 moles Al(OH)₃ x 3 moles HCl/mole Al(OH)₃ = 0.04659 moles

Concentration of HCl = moles/liter = 0.04659 moles/0.03518 L = 1.32 M (to 3 significant figures)

Answer 2

Step 1: Write out the acid base reaction and balance.

_HCl + _Al(OH)₃ --> _H₂O + _AlCl₃

Step 2: Calculate how many moles of Al(OH)₃ were neutralized by acid.

Molarity (mol/L) x Volume (L) = mol

Step 3: Use your balanced equation to figure out how many moles of HCl are in the volume used to neutralize the base.

The relationship between the number of moles of acid and moles of aluminum hydroxide will help you figure out how many moles of acid are contained in the volume given in the problem. For example (THIS IS NOT REPRESENTATIVE OF WHAT THE ACTUAL ANSWER IS), if you calculated that you have 0.002 mol Al(OH)3 and the molar ratio of HCl to Al(OH)3 is 2:1, then you would use multiplication as such: 0.02 mol Al(OH)3 (2 mol HCl / 1 mol Al(OH)3) = 0.04 mol HCl

Step 4: Divide the number of moles HCl by the volume of acid in liters to get molarity.