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What is the Ka for C3H6O3 in a titration with NaOH?
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[QUOTE="Quantum Defect, post: 5051014, member: 537816"] You have three bits of information here with which to figure this out: (1) 10 cm^3 of 7.2 g/dm^3 of monoprotic acid (2) Equivalence point is reached with addition of 15.73 cm^3 of 0,05 M NaOH (typo above) ==> moles OH- = moles C3H6O3 (3) pH at equiv. point is "8" [The pH in the problem statement must refer to the pH of the equivalence point. 0.05 M NaOH [B]does not[/B] have pH = 8 ] So, you have at the end, a solution of the conj. base of your acid. From the pH of this solution, you can find Kb, and from this you can find Ka of the cong. acid. [/QUOTE]
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What is the Ka for C3H6O3 in a titration with NaOH?
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