A student prepares a dilute solution of sodium hydroxidem, NaOH (aq), starting with 6 M sodium hydroxide. She then titrates a 1.372 g samle of KHP with the dilute sodium hydroxide solution, NaOH (aq), calculate the molar concentration of the sodium hydroxide solution, NAoH (aq)

M = 0.3077 M

Explanation:

As I said in the comments, you are missing the required volume of the base to react with the KHP. I found this on another site, and the volume it used was 21.84 mL.

Now, KHP is a compound often used to standarize NaOH or KOH solutions. This is because it contains a mole ratio of 1:1 with the base, so it's pretty easy to use and standarize any base.

Now, as we are using an acid base titration, the general expression to use when a acid base titration reach the equivalence point would be:

n₁ = n₂   (1)

This, of course, if the mole ratio is 1:1. In the case of KHP and NaOH it is.

Now, we also know that moles can be expressed like this:

n = M * V   (2)

And according to this, we are given the volume of base and the required mass of KHP. So, if we want to know the concentration of the base, we need to get the moles of the KHP, because in the equivalence point, these moles are the same moles of base.

The reported molar mass of KHP is 204.22 g/mol, so the moles are:

n = 1.372 / 204.22 = 6.72x10⁻³ moles

Now, we will use expression (2) to get the concentration of the diluted base:

n = M * V

M = n / V

M = 6.72x10⁻³ / 0.02184

M NaOH = 0.3077 M

This is the concentration of the dilute solution of NaOH