What is the molarity of an NaOH solution if 4.37 mL is titrated by 11.1 mL of 0.0904 M HNO3?

To determine the molarity of the NaOH solution, we first need to understand the stoichiometry of the neutralization reaction between NaOH and HNO3. The balanced chemical equation for this reaction is:

[ \text{NaOH (aq)} + \text{HNO}_3 \text{(aq)} \rightarrow \text{NaNO}_3 \text{(aq)} + \text{H}_2\text{O (l)} ]

From the equation, we see that one mole of NaOH reacts with one mole of HNO3.

Next, we calculate the number of moles of HNO3 used in the titration. We can use the formula:

[ \text{moles} = \text{molarity} \times \text{volume (L)} ]

Given that the molarity of HNO3 is 0.0904 M and the volume used is 11.1 mL (which we need to convert to liters by dividing by 1000), we perform the following calculation:

[ \text{moles of HNO3} = 0.0904 , \text{mol/L} \times (11.1 ,