During the standardization of NaOH, the mass of pure potassium biphthalate is 0.3001 g. The volume...

During the standardization of NaOH, the mass of pure potassium biphthalate is 0.3001 g. The volume of NaOH used is 13.54 mL. What is the molarity of the NaOH solution in M to 3 decimal places? Molecular weight NaOH: 39.997 Molecular weight potassium biphthalate: 204.263 g/mol

Expert Solution

KHC8H4O4 (aq) + NaOH(aq) -------------> NaKC8H4O4 + H2O

1 mole 1mole

no of moles of KHP = W/G.M.Wt

= 0.3001/204.263 = 0.00147 moles

1 mole of KHP react with 1 mole of NaOH

0.00147 moles of KHP react with 0.00147 moles of NaOH

no of moles of NaOH = molarity * volume in L

0.00147 = molarity * 0.01354

molarity = 0.00147/0.01354 = 0.109M

molarity of NaOH = 0.109M

queen_honey_blossom answered 2 years ago

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