A 1.0L buffer solution contains 0.100 mol of HC2H3O2 and 0.100 mol of NaC2H3O2. The value of Ka for HC2H3O2 is 1.8Ă-10â'5. Part A Part complete Calculate the pH of the solution upon the addition of 0.015 mol of NaOH to the original buffer. Express the pH to two decimal places. pH p H = 4.88 Previous Answers Correct In the buffer solution before the addition of a strong base, [Aâ']=[HA], so its pH is equal to the pKa of the weak acid. Adding a strong base to a buffer solution increases the concentration of Aâ' (C2H3O2â') and decreases the concentration of HA (HC2H3O2). Therefore, the final pH is slightly greater than the pKa for this weak acid. Significant Figures Feedback: Your answer 4.8760 was either rounded differently or used a different number of significant figures than required for this part. Part B Calculate the pH of the solution upon the addition of 10.0 mL of 1.00 M HCl to the original buffer. Express the pH to two decimal places. pH p H = nothing Request Answer

Question

Reagan Meyer

Chemistry

18 February, 15:26

A 1.0L buffer solution contains 0.100 mol of HC2H3O2 and 0.100 mol of NaC2H3O2. The value of Ka for HC2H3O2 is 1.8Ă-10â'5. Part A Part complete Calculate the pH of the solution upon the addition of 0.015 mol of NaOH to the original buffer. Express the pH to two decimal places. pH p H = 4.88 Previous Answers Correct In the buffer solution before the addition of a strong base, [Aâ']=[HA], so its pH is equal to the pKa of the weak acid. Adding a strong base to a buffer solution increases the concentration of Aâ' (C2H3O2â') and decreases the concentration of HA (HC2H3O2). Therefore, the final pH is slightly greater than the pKa for this weak acid. Significant Figures Feedback: Your answer 4.8760 was either rounded differently or used a different number of significant figures than required for this part. Part B Calculate the pH of the solution upon the addition of 10.0 mL of 1.00 M HCl to the original buffer. Express the pH to two decimal places. pH p H = nothing Request Answer

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Answers (1)

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Jakayla Santos · Answer 1

A: 4.88; B: 4.66

Step-by-step explanation:

Part A. pH on adding base

HC₂H₃O₂ + OH⁻ ⇌ H₂O + C₂H₃O₂⁻; Kₐ = 1.8 * 10⁻⁵

I/mol: 0.10 0.015 0.10

C/mol: - 0.015 - 0.015 + 0.015

E/mol: 0.085 0 0.115

pH = pKₐ + log{[salt]/[acid]}

pKₐ = - logKₐ

pKₐ = - log (1.8 * 10⁻⁵)

pKₐ = 4.74

pH = 4.74 + log (0.115/0.085)

pH = 4.74 + log (1.35)

pH = 4.74 + 0.131

pH = 4.88

Part B. pH on adding acid

Moles of acid added = 0.0100 L * (1.00 mol/1 L) = 0.0100 mol

HC₂H₃O₂ + H₂O ⇌ H₃O⁺ + C₂H₃O₂⁻

I/mol: 0.10 0.0100 0.10

C/mol: + 0.0100 - 0.0100 - 0.0100

E/mol: 0.11 0 0.09

pH = pKₐ + log{[salt]/[acid]}

pH = 4.74 + log (0.09/0.11)

pH = 4.74 + log (0.818)

pH = 4.74 - 0.087

pH = 4.66