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HA + NaOH → → NaA + H₂O Kelly split a 50.0 mL solution of weak acid into two separate 25.0 mL parts. She then added NaOH one of the parts until she observed the endpoint. After then added both parts, she measured the pH of the half-titrated weak acid solution as 8.70. What is the Ka for the acid? Ka = [?] x 10[?] Coefficient (green) Exponent (yellow) Enter

HA + NaOH → → NaA + H₂O Kelly split a 50.0 mL solution of weak acid into two separate 25.0 mL parts. She then added NaOH one of the parts until she observed the endpoint. After then added both parts, she measured the pH of the half-titrated weak acid solution as 8.70. What is the Ka for the acid? Ka = [?] x 10[?] Coefficient (green) Exponent (yellow) Enter

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter15: Additional Aqueous Equilibria
Section: Chapter Questions
Problem 93QRT: When 40.00 mL of a weak monoprotic acid solution is titrated with 0.100-M NaOH, the equivalence...
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HA + NaOH → → NaA + H₂O Kelly split a 50.0 mL solution of weak acid into two separate 25.0 mL parts. She then added NaOH one of the parts until she observed the endpoint. After then added both parts, she measured the pH of the half-titrated weak acid solution as 8.70. What is the Ka for the acid? K₂ = [? ] × 10¹²] a Coefficient (green) Exponent (yellow) Enter
Transcribed Image Text:HA + NaOH → → NaA + H₂O Kelly split a 50.0 mL solution of weak acid into two separate 25.0 mL parts. She then added NaOH one of the parts until she observed the endpoint. After then added both parts, she measured the pH of the half-titrated weak acid solution as 8.70. What is the Ka for the acid? K₂ = [? ] × 10¹²] a Coefficient (green) Exponent (yellow) Enter
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