1. A sample of potassium hydrogen phthalate (KHC 8 H 4 O 4 ), which is an acid,

Question

1. A sample of potassium hydrogen phthalate (KHC8H4O4), which is an acid, with a mass of 0.4304 g is dissolved in water and used to standardize a solution of potassium hydroxide. An endpoint is reached when 30.48 mL of potassium hydroxide has been added. Determine the molarity (in mol/L) of the potassium hydroxide solution. Report your answer to four significant figures.

2. The standardized potassium hydroxide solution is then used to titrate 18.79 mL of a solution of phosphoric acid (H3PO4). An endpoint is reached when 32.67 mL of potassium hydroxide has been added. Determine the molarity (in mol/L) of the phosphoric acid solution. Report your answer to four significant figures.

Explanation / Answer

1) KOH(aq.)+ KHC8H4O4(aq.) K2C8H4O4 + H2O

Molarity of KHC8H4O4 = number of moles of KHC8H4O4 /volume of solution in litre

Number of moles of KHC8H4O4 = Mass of KHC8H4O4 (g) / Molecular mass of KHC8H4O4

                                                = 0.4304/ 204.23

= 0.0021 moles

At equivalence point the # moles OH- = # moles of H+ (As ratio of acid base in balanced equation is 1:1)

# moles of KOH = 0.0021 moles

Molarity of KOH = number of moles of KOH /volume of solution in litre

= 0.0021/ 0.03048

= 0.06914 M

2) M1V1 (KOH) = M2V2 (H3PO4)

0.06914 X 32.67 = M2 X 18.79

M2 = (0.06914 X 32.67) / 18.79

= 0.1202 M

Molarity of (H3PO4) is ~ 0.1202 M.

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