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Test: Buffer solutions (August 1) - NEET MCQ


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10 Questions MCQ Test - Test: Buffer solutions (August 1)

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Test: Buffer solutions (August 1) - Question 1

Which of the following is not an acidic buffer?

Detailed Solution for Test: Buffer solutions (August 1) - Question 1
An acidic buffer has a pH value of less than 7, Acetic Acid-Sodium Acetate and boric acid-borax are examples of acidic buffers, but ammonium Hydroxide-ammonium chloride has a pH of greater than 7, so they are basic buffers.
Test: Buffer solutions (August 1) - Question 2

Buffer capacity of a buffer is given as two units for a change in pH by Unity. Then what is the number of moles of acid or base, added in one litre of the solution?

Detailed Solution for Test: Buffer solutions (August 1) - Question 2
Buffer capacity is denoted by ? = number of moles of acid or base added to one litre of the buffer by a change in pH. Here the change in pH is given as 1, and the buffer capacity is given as 2, therefore, by substituting, we get that 2 moles of acid or base are added in one litre of the solution.
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Test: Buffer solutions (August 1) - Question 3

Buffer solution is destroyed when _____________

Detailed Solution for Test: Buffer solutions (August 1) - Question 3

If the addition of a strong acid or base changes the pH of a buffer by one unit, the buffer solution is assumed to be destroyed, meaning the new pH = pKa ± 1; that means [salt]/[acid] or [acid]/[salt] = 10 or 1/10.

Test: Buffer solutions (August 1) - Question 4
What is the buffer capacity if 3 moles are added in 5 litres of the solution to change the pH by 2 units?
Detailed Solution for Test: Buffer solutions (August 1) - Question 4
Buffer capacity is defined as the number of moles of acid or base added in one litre of the solution to change the pH by Unity. Therefore, here buffer capacity = 3/5 divided by 2 = 0.6/2 = 0.3. The buffer capacity is given as 0.3.
Test: Buffer solutions (August 1) - Question 5
Which of the following is an equation used to calculate the pH of a buffer solution for an acidic buffer?
Detailed Solution for Test: Buffer solutions (August 1) - Question 5
Equation that is used to calculate the pH of a buffer solution for an acidic buffer is pH = pKa + log[salt]/[acid]. This equation is known as the Henderson-Hasselbalch equation; it is used for making buffer solutions.
Test: Buffer solutions (August 1) - Question 6
If 0.20 mol/L CH3COOH and 0.50 mol/L CH3COO– together make a buffer solution, calculate the pH of the solution if the acid dissociation constant of CH3COOH is 1.8 × 10-5.
Detailed Solution for Test: Buffer solutions (August 1) - Question 6
We have the Henderson-Hasselbalch equation as pH = pKa + log[salt]/[acid]. So by substituting the concentrations of salt and acid along with the acid dissociation constant, we get pH = -log[1.8 × 10-5] + log [0.50 mol/L]/[0.20 mol/L] = 5.14.
Test: Buffer solutions (August 1) - Question 7
Note that the pKa here is given by 4.752, a buffer is made using 0.8 M acetic acid and 1 M Sodium Acetate what do you think its pH is (log10/8 = 0.097)?
Detailed Solution for Test: Buffer solutions (August 1) - Question 7
According to the Henderson-Hasselbalch equation pH = pKa + log[salt]/[acid], if we substitute the concentration of salt as 1 M and the concentration of acid as 0.8 M, pH = 4.752 + 0.097 = 4.849, which is approximately 4.85.
Test: Buffer solutions (August 1) - Question 8
If the pH of a substance is given by 3 then what is the pOH of the substance?
Detailed Solution for Test: Buffer solutions (August 1) - Question 8
We know that the sum of the pH and pOH of any substance is equal to 14, that is pH + pOH = 14. So here the pH of a substance is given as 3; the pOH of the substance = 14 – 3 = 11.
Test: Buffer solutions (August 1) - Question 9
Which of the following do you think is a correct statement?
Detailed Solution for Test: Buffer solutions (August 1) - Question 9
Ammonium hydroxide / ammonium chloride is a basic buffer, Henderson-Hasselbalch equation is given by pH = pKa + log[salt]/[acid], and pH + pOH = 14. So the only correct statement is that boric acid / borax is an acidic buffer.
Test: Buffer solutions (August 1) - Question 10

Which will make basic buffer?

Detailed Solution for Test: Buffer solutions (August 1) - Question 10

Concept:

Basic buffer solution:

  • As the name suggests, a basic buffer maintains the basic pH in the solution.
  • It is also called an alkaline buffer solution.
  • It is prepared by mixing a weak base with a salt of strong acid.
  • These buffers have a pH > 7.
  • For example - a buffer of NH4OH and NH4Cl.

Explanation:

The basic buffer solution is made by mixing of a weak base with strong acid in such a way to maintain the pH of the solution basic or >7.

⇒ So, a basic buffer has a weak base and a salt of strong acid in the solution. 

⇒In 1st solution we have - 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH3COOH

total moles of NaOH initially 50 × 0.1 = 5 mmol
total moles of CH3COOH initially 25 × 0.1 = 2.5 mmol
2.5 mmole of NaOH neutralize the 2.5 mmol of CH3COOH and formed 2.5 mmole of sodium salt of acetic acid.
2.5 mmol of NaOH remains in the solution.
So, the solution has 2.5 mmol NaOH + 2.5 mmol CH3COONa, which is only a basic solution as NaOH is a strong base, not a weak base.
 ⇒ In 2nd solution we have - 100 mL of 0.1 M CH3COOH + 100 mL of 0.1 M NaOH

10 mmol of CH3COOH and 10 mmol of NaOH neutralize each other.
Same amount of acid, base, and salt will not form a basic buffer.
⇒ In the 3rd solution we have - 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH4OH

10 mmol of HCl reacts with only 10 mmol of NH4OH and forms 10 mmol of NH4Cl.
10 mmol of NH4OH remains in the solution.
Hence, the solution will have 10 mmol of NH4OH + 10 mmol of NH4Cl. 
It makes a basic buffer solution, as the solution has a weak base with a salt of strong acid.
⇒ 4th solution - 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH

It is a neutral solution as the same amount of acid and base neutralize each other.

Hence only, option c (100 mL of 0.1 M HCl + 200 mL of 0.1 M NH4OH) makes a basic buffer solution.

∴ the correct answer is option C.

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