chem help on both problems please A 0 397-g sample of potassium hydrogen phthala

Question

chem help on both problems please A 0 397-g sample of potassium hydrogen phthalate. KHC8H4O4 (molar mass=204.44g/mol) is dissolved with 50 mL of deionized water in a 125 ml.Erlenmeyer flask The sample is titrated to the phenolphthalein endpoint with 16.22 mL of a (sodium hydroxide solution What is the molar concentration of the NaOH solution? (Show all calculations) A 25.00-mLaliquot of a nitric acid solution of unknown concentration is pipetted into a 125-mL Erlenmeyer flask and 2 drops of phenolphthalein are added The above sodium hydroxide solution (the tirrant) is used to titrate the nitric acid solution (the analyte) If 12.75 mL of the titrant is dispensed from a buret in causing a color change of the phenolphthalein. what is the molar concentration of the nitric acid solution?

Explanation / Answer

Solution :-

Q5 a) 0.397 g KHP

Lets first calculate the moles of the KHP

Moles of the KHP = mass of KHP / molar mass

                                 = 0.397 g / 204.44 g per mol

                                 = 0.001942 mol

Now using the mole ratio of the KHP and NaOH lets calculate the moles of NaOH

Mole ratio of the KHP and NaOH is 1 :1 therefore the moles of the NaOH reacted are same as moles of KHP

Therefore moles of NaOH = 0.001942 moles

Now lets calculate the molarity of the NaOH using the moles and volume

Volume of NaOH used = 16.22 ml = 0.01622 L

Molarity of NaOH = moles / volume in liter

                              = 0.001942 mol / 0.01622 L

                               = 0.120 M

Therefore the NaOH molarity = 0.120 M

Q5 b) molarity of the NaOH in the part a = 0.120 M volume of NaOH used = 12.75 ml

Volume of HNO3 = 25.0 ml

Molarity of the HNO3 = ?

Mole ratio of the HNO3 and NaOH is 1 :1

Therefore we canuse the following set up to calculate the molarity of the HNO3

Molarity of HNO3 = molarity of NaOH * volume of NaOH / volume of HNO3

                                 = 0.120 M * 12.75 ml / 25 ml

                                 = 0.0612 M

Therefore the molarity of the HNO3 = 0.0612 M

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