- 5 A 0 682 Gram Sample Of An Unknown Weak Monoprotic Acid Ha Was Dissolved In Water To Make 50 00 Ml Of Solution And W 1 (24.24 KiB) Viewed 56 times
5. A 0.682 gram sample of an unknown weak monoprotic acid, HA, was dissolved in water to make 50.00 ml of solution and w
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5. A 0.682 gram sample of an unknown weak monoprotic acid, HA, was dissolved in water to make 50.00 ml of solution and w
5. A 0.682 gram sample of an unknown weak monoprotic acid, HA, was dissolved in water to make 50.00 ml of solution and was titrated with a 0.135 M NaOH solution. After the addition of 10.6 milliliters of base, a pH of 5.65 was recorded. The equivalence point (end point) of the titration was reached after the addition of 27.4 milliliters of the 0.135 M 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 (H') at pH 5.65. (e) Calculate the value of the ionization constant, Ka of the acid HA.
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