Question

Initially a NaOH solution was standardized by titration with a sample of potassium hydrogenphthalate, KHC8H4O4, a monoprotic acid often used as a primary standard. A sample of pureKHC8H4O4 weighing 1.518 grams was dissolved in water and titrated with the NaOH solution. Toreach the equivalence point, 26.90 milliliters of base was required. Calculate the molarity of theNaOH solution. (Molecular weight: KHC8H4O4 = 204.2)

Answer 1

Answer:

Molarity of the NaOH solution is 0.2764 M.

Explanation:

Mass of potassium hydrogenphthalate = 1.518 g

Moles of potassium hydrogenphthalate =$\frac{1.518 g}{204.2 g/mol}=0.007434 mol$

$KHC_8H_4O_4+NaOH\rightarrow NaKC_8H_4O_4+H_2O$

According to reaction, 1 mole of potassium hydrogenphthalate reacts with 1 mole of NaOH, then 0.007434 moles pf potassium hydrogenphthalate will :

$\frac{1}{1}\times 0.007434 mol=0.007434$ of NaOH

Moles of NaOH = 0.007434 mol

Volume of NaOH solution =  26.90 mL = 0.02690 L

1 mL = 0.001 L

$Molarity=\frac{Moles}{Volume (L)}$

Molarity of the NaOH solution :

$=\frac{0.007434 mol}{0.02690 L}=0.2764 M$