A solution of phenol (HC,H;0) is prepared by dissolving 0.385 g of phenolin 2.00 L of H20. The solution has a pH = 6.29. Determine the value of Kafor phenol.( PREV12Based on your ICE table and definition of Ka, set up the expression for Ka and thenevaluate it. Do not combine or simplify terms.Ka =%3D5 RESET[0][0.385][6.29][4.09 x 10-][2.05 x 10][5.1 x 10-7][0.799]1.3 x 10-107.8 x 10°4.0 x 1032.5 x 10-4 A solution of phenol (HC6H;O) is prepared by dissolving 0.385 g of phenolin 2.00 L of H20. The solution has a pH = 6.29. Determine the value of Kafor phenol.12NEXT >Based on the given values, fill in the ICE table to determine concentrations of allreactants and products.HC6H;0(aq) +H2O(1)H3O*(aq)C6H5O¯(aq)+Initial (M)Change (M)Equilibrium(M)5 RESET0.3856.29-6.294.09 x 103-4.09 x 10-32.05 x 10-3-2.05 x 1035.1 x 10-7-5.1 x 10-70.799-0.799

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A solution of phenol (HC6H50) is prepared by dissolving 0.385 g of phenol in 2.00 L of H2O. The solution has a pH = 6.29. Determine the value of Ka for phenol. ( PREV 1 2 Based on your ICE table and definition of Ka, set up the expression for Ka and then evaluate it. Do not combine or simplify terms. Ka 5 RESET [0] [0.385] [6.29] [4.09 x 10] [2.05 x 10] [5.1 x 10-7] [0.799] 1.3 x 10-10 7.8 x 10° 4.0 x 103 2.5 x 10-4

A solution of phenol (HC6H50) is prepared by dissolving 0.385 g of phenol in 2.00 L of H2O. The solution has a pH = 6.29. Determine the value of Ka for phenol. ( PREV 1 2 Based on your ICE table and definition of Ka, set up the expression for Ka and then evaluate it. Do not combine or simplify terms. Ka 5 RESET [0] [0.385] [6.29] [4.09 x 10] [2.05 x 10] [5.1 x 10-7] [0.799] 1.3 x 10-10 7.8 x 10° 4.0 x 103 2.5 x 10-4

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter14: Acids And Bases

Section: Chapter Questions

Problem 30QRT

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