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682 g sample of an unknown weak monoprotic organic acid, HA was dissolved in sufficient water to make 50 milliliters of solution and was titrated...

This question was answered on Mar 17, 2010. View the Answer
A 0.682 g sample of an unknown weak monoprotic organic acid, HA was dissolved in sufficient water to make 50 milliliters of solution and was titrated with a 0.135 molar NaOH solution. After the addition of 10.6 milliliters of base, a pH of 5.65 was recorded. The equivalence point (end point) was reached after the addition of 27.4 milliliters of the 0.135 molar NaOH.
a. Calculate the number of moles of acid in the original sample.
b. Calculate the molecular weight of the acid HA.
c. Calculate the number of moles of unreacted HA remaining in solution when the pH was 5.65
d. Calculate the [H3O+] at pH =5.65
e. Calculate the value of the ionization constant, Ka, of the acid HA

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This question was asked on Mar 17, 2010 and answered on Mar 17, 2010.

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