Answer to Question #198750 in Inorganic Chemistry for Rhowinne Marie Bar

Question

a titration of potassium hydrogen phthalate required 38.14 mL of NaOH solution to reach an end point detected by phenolphthalen indicator. Find the concentration of NaOH. Weight of potassium and phthalate is 08.215 g.

Answer 1

Potassium hydrogen pathalate (KHP ) can provide 1 mol of H+ ion , and NaOH (sodium hydroxide) can provide 1 mol of OH- ion per mol. They react 1 mol to 1 mol.

so moles of Potassium hydrogen pathalate required =moles of NaOH

so we calculate moles of KHP first,

$n(moles)= {G.M \over M.M}$

$$={ 08.215 g \over 204.22}\\$$

$$=0.04023$$

no equating these moles for NaOH.

we find concentration of NaOH(38.14 mL)

$$Molarity={n\over V}$$

$$={0.04023\over 0.03814}$$

$$=1.0548 M$$