pKa = pH – log([A¯]/[HA])pKp = pOH – log([BH+]/[B])Notice that for pKp a conversion from pH to pOH using the equation pH = 14 – pOH will be necessary. Use this procedure to determine the pH, pKa, and pKp values for the following solutions.Match the words to the appropriate blanks in the sentences below. Make certain each sentence is complete before submitting your answer.• View Available Hint(s)ResetHelp7.501. An acidic solution with an initial acid concentration of 0.010 mol/L and an equilibrium1.99concentration for A- of 1.79 × 10¬has a pH of. The pKa of the acid is1.584.752. An acidic solution with an initial acid concentration of 0.015 mol/L and an equilibriumconcentration for A- of 1.02 × 10-2 has a pH ofThe pKa of the acid is11.573.433. A basic solution with an initial base concentration of 0.021 mol/L and an equilibrium1.66concentration for BH+ of 1.38 × 10¬2 has a pH of. The pKb of the base is12.144. A basic solution with an initial base concentration of 0.041 mol/L and an equilibriumconcentration for BH† of 3.70 × 10–3has a pH ofThe pKp of the base is

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pKa = pH – log([A¯]/[HA]) pKb = pOH – log((BH+]/[B]) Notice that for pKp a conversion from pH to pOH using the equation pH = 14 – pOH will be necessary. Use this procedure to determine the pH, pKa, and pK, values for the following solutions. Match the words to the appropriate blanks in the sentences below. Make certain each sentence is complete before submitting your answer. > View Available Hint(s) Reset Help 7.50 1. An acidic solution with an initial acid concentration of 0.010 mol/L and an equilibrium concentration for A- of 1.79 x 10-5 has a pH of The pK, of the acid is 1.99 1.58 4.75 2. An acidic solution with an initial acid concentration of 0.015 mol/L and an equilibrium concentration for A of 1.02 x 10-2 has a pH of The pK, of the acid is 11.57 3.43 3. A basic solution with an initial base concentration of 0.021 mol/L and an equilibrium 1.66 concentration for BH+ of 1.38 x 10-2 has a pH of The pKp of the base is 12.14

pKa = pH – log([A¯]/[HA]) pKb = pOH – log((BH+]/[B]) Notice that for pKp a conversion from pH to pOH using the equation pH = 14 – pOH will be necessary. Use this procedure to determine the pH, pKa, and pK, values for the following solutions. Match the words to the appropriate blanks in the sentences below. Make certain each sentence is complete before submitting your answer. > View Available Hint(s) Reset Help 7.50 1. An acidic solution with an initial acid concentration of 0.010 mol/L and an equilibrium concentration for A- of 1.79 x 10-5 has a pH of The pK, of the acid is 1.99 1.58 4.75 2. An acidic solution with an initial acid concentration of 0.015 mol/L and an equilibrium concentration for A of 1.02 x 10-2 has a pH of The pK, of the acid is 11.57 3.43 3. A basic solution with an initial base concentration of 0.021 mol/L and an equilibrium 1.66 concentration for BH+ of 1.38 x 10-2 has a pH of The pKp of the base is 12.14

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

Section14.8: Acid-base Reactions Of Salts

Problem 14.21CE

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