IIT JAM Exam  >  IIT JAM Questions  >  40 millimoles of NaOH are added to 100 ml of ... Start Learning for Free
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?
Most Upvoted Answer
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.
Explore Courses for IIT JAM exam
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?
Question Description
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?.
Solutions 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? in English & in Hindi are available as part of our courses for IIT JAM. Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of 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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an ample number of questions to practice 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? tests, examples and also practice IIT JAM tests.
Explore Courses for IIT JAM exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev