What is the molarity of a NaOH solution if 38.5 mL are needed to titrate a 0.7254 sample of KHP (204.2 g/mol)?

Question

Can anyone help me solve my chemistry homework problem? If so, that would be so much help!! :) Thanks!

Answer 1

I guess it is 0.7254 g

KHP is Potassium Hydrogen Phthalate

NaOH + KHP -> KNaP+ H2O

The stoichiometry is 1:1, so moles NaOH=moles KHP
but moles=M*V and also moles= mass/MW, so

M(NaOH)*V(NaOH)= mass(KHP) / MW(KHP)

and

M(NaOH) = mass(KHP) / (MW(KHP) * V(NaOH))

since M is mole/L you need to convert mL into L.

Answer 2

What is KHP? There is nothing by that formula that has a molar mass of 204.2g/mol.

Is it 0.7254ml ? 0.7254M? 0.7254N? 0.7254g?

Not enough info is provided.

But either way you will need to find out the number of mole equivalents in the titrated acid and divide that my the number of liters of NaOH to get the molarity. This requires a balanced equation representing the acid base reaction if it isn't a 1:1 acid:base ratio.

Molarity= #moles/#liters

Answer 3

molarity of solution 1 x volume of solution 1 = molarity of solution 2 x volume of solution 2

moles of solution1= moles of solution 2

.7254g/204.2g/mol = x moles

x moles/ y mLs = molarity.

I believe NaOH and KHP (potassium hydrogen phthalate) react one to one.

You also need to convert mL to liters because molarity is expressed in g/liter