Question

How many milliliters of 0.100 M HNO3 are needed to neutralize the following solutions?
A. 45.0 mL of 0.667 M KOH
B. 58.5 mL of 0.0100 M Al(OH)3

Answer 1

Writ the equations and balance them.
HNO3 + KOH ==> KNO3 + H2O
3HNO3 + Al(OH)3 ==> 3H2O + Al(NO3)3

millimoles KOH = mL x M = 45.0*0.667 = 30.015
Convert mmoles KOH to mmoles HNO3 using the coefficients in the balanced equation. That is done as
mmoles HNO3 = mmoles KOH x (1 mole HNO3/1 mole KOH) = 30.015 x (1/1) = 30.015 mmoles HNO3.
Then M = mmoles/mL and solve for mL = mmoles/M = 30.015/0.100 = 300.15 mL. Round to the correct number of sig figures.
The second part is done the same way but the ratio of acid/base is not 1:1 as it is with KOH/HNO3.