Answer:
6 ml of 0.50 M NaOH
Step-by-step explanation:
We need to start with a balanced equation. For this reaction, it is simple:
NaOH + HNO3 = NaNO3 + H2O
The balanced equation tells us we need 1 mole of NaOH to completely neutralize (titrate) 1 mole of HNO3. That means a molar ratio (NaOH/HNO3) of 1:1. We need the same number of moles of NaOH as we have of HNO3.
Moles HNO3
If we can determine the number of moles of HNO3 there are in 15.0 ml of 0.20M HNo3 solution, then that will also be the number of moles of NaOH that are required for titration.
Always remember that M means "moles/liter." It always helps to use the full unit in doing these calculations, because it provides guidelines as to what conversions and order of operations are needed.
15.0 ml of 0.20 moles/liter HNO3 [M is replaced with moles/liter]
Note: Volume(V) x Concentration(C) = Moles
We can see that if we multiply these two numbers, that the volume term will cancel, once we convert ml into liters. [1 liter/1000 ml]
V x C = moles
(0.20 moles/liter)*(15.0 ml) = moles HNO3
(0.20 moles/liter)*(15.0 ml)*(1 liter/1000 ml) [Convert L to ml]
Cancel units: ml cancels, and then liters also cancels, leaving us with just moles, which is our goal.
There are 0.003 moles of HNO3 in 15 ml of 0.20M HNO3.
We know from the balanced equation that we need the same moles of NaOH for titration (neutralization). So the question then becomes how many ml of 0.50M NaOH will provide 0.003 moles of NaOH.
Volume x Concentration = Moles Volume, V, is the unknown.
V*(0.50 moles/liter) = 0.003 moles NaOH
V = 0.006 liters
Or, volume = 6 ml.