Acids and Bases

Question

What volume of a 2 % NaOH solution is required to neutralize 40 mL of 8 % HCl? Provide your answer to one decimal place only. [MW of HCl = 36.5 g/mol, MW of NaOH = 40.0 g/mol]

Answer 1

I am assuming that the percent compositions are mass / volume. So, 8% HCl would have 8 g HCl per 100 mL.

Since there are 40 mL of HCl, the total grams of HCl is 8 g / 100mL x 40 mL = 3.2 g. We now need to convert this mass to moles which we do by dividing 3.2 g / 36.g/mole = 8.77 x 10^-2 moles HCl.

We can express the moles of NaOH in a similar way. 2 g/100 mL x V / 40 g/mole = 5 x 10^-4 x V moles, where V is the unknown volume of NaOH required to neutralize the HCl.

Since the neutralization reaction in this case is mole to mole, i.e. HCl + NaOH ==> NaCl + H2O, we can equate the two expressions and solve for V.

5 x 10^-4 x V moles = 8.77 x 10^-2 moles

V = 175.4 mL of NaOH required.

Verify my math.