Question #3cfec

5 ml
Each mole of NaOH neutralizes 1 mole of acid. The 0.2 M slution of HNO3 contains 0.02 moles in 100 ml.The 4 M NaOH solution contains 4 millimoles per milliter. Therefore 5 ml of NaOH solution are required to neutralize the acid.

When the values given to you allow you to do this quick calculation, here's how you should approach this problem.

You know that you have a #1:1# mole ratio between nitric cid and sodium hydroxide, which means that, in order to have a complete neutralization, you need equal numbers of moles of each compound.

Now, notice that the sodium hydroxide solution is 20 times more concentrated than the nitric acid solution. This means that, for a given volume, the sodium hydroxide solution will contain 20 times more moles of #NaOH# than the nitric acid solution will contain of #HNO_3#.

Therefore, if you use a volume of sodium hydroxide solution 20 times smaller than the volume of nitric acid solution, you'll get equal numbers of moles of acid and of base.

In your case, if you have 100.0 mL of nitric acid, the volume of the sodium hydroxide solution will be

#V_(NaOH) = "100.0 mL"/20 = "5.0 mL"#

More often than not, the values given to you will not allow for such quick calculations.