titration problem

Question

A student weighs out 0.568 g of KHP (molar mass = 204 g/mol) and titrates to the equivalence point with 36.78 mL of a stock NaOH solution. What is the concentration of the stock NaOH solution? KHP is an acid with one acidic proton.

Answer 1

Ok,KHP is a monoprotic acid. In a titration where neutralization occurs, it is 1 H⁺ to 1 OH^- molar ratio. If we can figure out how many moles of KHP there is, we would find how many moles of H⁺ there is ( KHP to H⁺ is a 1 to 1 molar ratio-monoprotic acid thing again). This can be found by dividing the molar mass of KHP into the mass of KHP (.568/204-the mass,gram units, cancel and moles remain if you use the dimensional analysis method).

You get .00278 moles of KHP. Now let's apply the molar ratio logic to obtain the moles Of OH^-: .00278 moles of KHP means there's .00278 moles of H⁺, .00278 moles of H+ means the neutralized solution has.00278 moles of OH-. NaOH is a base with a 1 OH- to 1 NaOH molar ratio so that there are.00278 moles of NaOH. To find molarity (concentration = mol/L), convert the 36.78ml of NaOH into L, and divide this into .00278 moles of NaOH...

Answer 2

Write a correctly balanced equation for the reaction taking place. One must assume that the KHP referred to is potassium hydrogen phthalate, and not potassium hydrogen phosphate, otherwise the molar mass would be incorrect. That's one problem with abbreviations. At any rate...

NaOH + C₈H₅KO₄ ==> C₈H₄KNaO₄ + H₂O

moles acid present = 0.568 g x 1 mol/204 g = 0.002784 moles acid

moles NaOH needed to neutralize = 0.002784 moles NaOH since balanced equation shows 1:1 mole ratio

Molarity of the NaOH = moles/liter = 0.002784 moles/0.03678 L = 0.07570 M = 0.0757 M (3 sig. figs.)