What quantity of 18.68 M KOH solution is needed to neutralize 0.2841 grams of KHP? What is the concentration of the KOH solution?

KHP stands for potassium hydrogen phthalate which is the monopotassium salt of phthalic acid.

KHP dissociates completely in water to form K (+) + HP (-); and HP (-).

HP (-) dissolves as per HP (-) + H2O ⇄ P (2-) + H3O (+).

That means that KHP the neutralization ratio of KOH and KPH is 1:1.

So, 1 mole of KOH neutralizes 1 mol of KPH.

Calculate the moles of KPH in 0.2841 grams.

You need the molar mass of KPH. Its molecular formula is C8H5KO4 and its molar mass is 204.22 g/mol

= n = 0.2841 g / 204.22 g/mol = 0.00139 moles of KPH.

Now, calculate the volumen of 18.68 M KOH solution that contains that number of moles:

M = n / V = > V = n / M = 0.00139 moles / 18.68 M = 0.0000744 l = 0.0744 ml.

The concentration of the KOH solution is given in the statement: 18.68 M.

That is a weird value for the concentration so may be your question has some mistake.

Answer: 0.0744 ml