Potassium hydrogen phthalate (molar mass = 204.2g/mol) is used to standardize sodium hydroxide. If 26.37 mL of NaOH(aq) is required to titrate 0.7719g KHP to the equivalence point, what is the concentration of the NaOH(aq)

(26.37 mL NaOH)(x) = (1 mol KHP/.7719g KHP)(204.2g KHP/mol KHP)

10.03M NaOH

I think it would be easier to do this in steps. I think you are confusing yourself in trying to do it all in one step.

mols KHP = 0.7719 g x (1 mol/204.2 g) = ?? (which is the inverse of what you have written. Look at the units for the right side of what you wrote. mols cancels with mols and g cancels with g so the right hand side of your equation, which should come out in moles, is unitless).
mols NaOH = same thing (that's when the indicator tells you that mols of one = moles of the other---to be more exact when the equivalents of one = equivalents of the other).
mols NaOH = M x L.
You know mols NaOH and L, calculate M.
I don't get anything close to 10 M.

.0143 mol

To determine the concentration of NaOH(aq), we can use the equation:

(26.37 mL NaOH)(x) = (1 mol KHP/0.7719g KHP)(204.2g KHP/mol KHP)

First, calculate the moles of KHP used:

moles KHP = 0.7719g KHP / 204.2g/mol KHP

moles KHP ≈ 0.003779 mol KHP

Now, based on the balanced chemical equation for the reaction between KHP and NaOH, we know that 1 mole of KHP reacts with 1 mole of NaOH:

1 mol KHP = 1 mol NaOH

Therefore, the moles of NaOH used is also approximately 0.003779 mol.

Now, we can calculate the concentration of NaOH:

(26.37 mL NaOH)(x) = 0.003779 mol NaOH

x ≈ 0.003779 mol NaOH / 26.37 mL

x ≈ 0.1434 mol/L

Thus, the concentration of NaOH(aq) is approximately 0.1434 M.

To determine the concentration of NaOH(aq), we can use the method of stoichiometry.

First, we need to calculate the number of moles of KHP (Potassium Hydrogen Phthalate) used in the titration. This can be done using the given molar mass of KHP (204.2 g/mol) and the mass of KHP used (0.7719 g).

Number of moles of KHP = Mass of KHP / Molar mass of KHP
= 0.7719 g / 204.2 g/mol
= 0.003776 mol

According to the balanced chemical equation between NaOH and KHP, the ratio of moles of NaOH to KHP is 1:1.
So, the number of moles of NaOH used is also 0.003776 mol.

Next, we convert the volume of NaOH used (26.37 mL) into liters.

Volume of NaOH used = 26.37 mL * (1 L / 1000 mL)
= 0.02637 L

Now, we can calculate the concentration of NaOH using the formula:

Concentration (Molarity) = Number of Moles / Volume (in liters)

Concentration of NaOH = 0.003776 mol / 0.02637 L
= 0.1430 M

Therefore, the concentration of NaOH(aq) is 0.1430 M.