Ascorbic acid (HC6H706) is a weak monoprotic acid (HA); A) Calculate the pH of a solution produced by mixing 100.00 mL of 0.0500 M ascorbic acid (HA) with 15.2 ml of 0.2500 M NaOH. Ka for ascorbic acid = 5 x10-5. (b) B) Calculate the pH after addition of 30.00- mL of 0.2500 M NaOH to the Ascorbic acid above. For part B, it states that to get the OH- it would be 0.0025 mol/ 0.13 L, but I do not understand where the 0.13 L comes from. May this please be explained? Thank you.
Ascorbic acid (HC6H706) is a weak monoprotic acid (HA); A) Calculate the pH of a solution produced by mixing 100.00 mL of 0.0500 M ascorbic acid (HA) with 15.2 ml of 0.2500 M NaOH. Ka for ascorbic acid = 5 x10-5. (b) B) Calculate the pH after addition of 30.00- mL of 0.2500 M NaOH to the Ascorbic acid above. For part B, it states that to get the OH- it would be 0.0025 mol/ 0.13 L, but I do not understand where the 0.13 L comes from. May this please be explained? Thank you.
Chapter15: Acid-base Equilibria
Section: Chapter Questions
Problem 8ALQ: You have a solution of the weak acid HA and add some of the salt NaA to it. What are the major...
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