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40 millimoles of NaOH are added to 100 ml of a 1.2 M HA and Y M NaA buffer resulting in a solution of pH 5.30. Assuming that the volume of the buffer remains unchanged, the pH of the buffer (K = 1.00 x 10'-5) is?
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40 millimoles of NaOH are added to 100 ml of a 1.2 M HA and Y M NaA bu...
Calculation of Initial Concentration of HA and NaA:

  • Number of moles of NaOH = (40 mmol/L) x (0.1 L) = 4 mmol

  • Number of moles of HA initially present = (1.2 mol/L) x (0.1 L) = 0.12 mol

  • Number of moles of NaA initially present = Y x (0.1 L)



Calculation of Concentration of HA and NaA after addition of NaOH:

  • 4 mmol of NaOH reacts with 4 mmol of HA to form A-

  • Remaining moles of HA = 0.12 - 4 = 0.08 mol

  • Remaining moles of NaA = Y - 4

  • Concentration of HA = 0.08 mol / 0.1 L = 0.8 M

  • Concentration of NaA = (Y - 4) mol / 0.1 L = (Y - 40) mmol / L



Calculation of pOH:

  • pOH = -log(OH-) = -log(40 mmol / 100 mL) = 1.40



Calculation of pH:

  • pH + pOH = 14

  • pH = 14 - pOH = 14 - 1.40 = 12.60



Calculation of [H+]:

  • Kw = [H+][OH-] = 1.00 x 10^-14

  • [H+] = Kw / [OH-] = 1.00 x 10^-14 / (40 mmol / 100 mL) = 2.50 x 10^-12 M



Calculation of Ka:

  • pH = pKa + log([A-] / [HA])

  • 12.60 = pKa + log(Y - 40 / 0.8)

  • pKa = 12.60 - log(Y - 40 / 0.8)



Calculation of [HA] at pH 5.30:

  • pH = pKa + log([A-] / [HA])

  • 5.30 = pKa + log(Y - 40 / [HA])

  • [HA] = (Y - 40) / (10^(5.30 - pKa))



Calculation of Y:

  • [HA] + [A-] = 0.1 L x Y M = Y / 10

  • [HA] = 0.8 M

  • [A-] = (Y - 40) / 10^pH - pKa

  • Y / 10 = 0.8 + (Y - 40) / 10^(5.
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40 millimoles of NaOH are added to 100 ml of a 1.2 M HA and Y M NaA buffer resulting in a solution of pH 5.30. Assuming that the volume of the buffer remains unchanged, the pH of the buffer (K = 1.00 x 10'-5) is?
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40 millimoles of NaOH are added to 100 ml of a 1.2 M HA and Y M NaA buffer resulting in a solution of pH 5.30. Assuming that the volume of the buffer remains unchanged, the pH of the buffer (K = 1.00 x 10'-5) is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about 40 millimoles of NaOH are added to 100 ml of a 1.2 M HA and Y M NaA buffer resulting in a solution of pH 5.30. Assuming that the volume of the buffer remains unchanged, the pH of the buffer (K = 1.00 x 10'-5) is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 40 millimoles of NaOH are added to 100 ml of a 1.2 M HA and Y M NaA buffer resulting in a solution of pH 5.30. Assuming that the volume of the buffer remains unchanged, the pH of the buffer (K = 1.00 x 10'-5) is?.
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