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.

0.0744 ml