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

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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)
Reset
Help
7.50
1. An acidic solution with an initial acid concentration of 0.010 mol/L and an equilibrium
1.99
concentration for A- of 1.79 × 10¬
has a pH of
. The pKa of the acid is
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 × 10-2 has a pH of
The pKa 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 × 10¬2 has a pH of
. The pKb of the base is
12.14
4. A basic solution with an initial base concentration of 0.041 mol/L and an equilibrium
concentration for BH† of 3.70 × 10–3
has a pH of
The pKp of the base is
Transcribed Image Text: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) Reset Help 7.50 1. An acidic solution with an initial acid concentration of 0.010 mol/L and an equilibrium 1.99 concentration for A- of 1.79 × 10¬ has a pH of . The pKa of the acid is 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 × 10-2 has a pH of The pKa 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 × 10¬2 has a pH of . The pKb of the base is 12.14 4. A basic solution with an initial base concentration of 0.041 mol/L and an equilibrium concentration for BH† of 3.70 × 10–3 has a pH of The pKp of the base is
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