# buffer solution practice problems with answers

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Questions: 15 | Attempts: 649  Settings  How good are you at solving buffer solution practice problems? Here, we have a few questions with answers based on the buffer solution in the quiz below. Give it a try and check how good your problem-solving skills are. A buffer solution is an aqueous solution consisting of a weak acid and its conjugate base. The quiz below will ask you to calculate the pH of chemical compounds using the Henderson–Hasselbalch equation. Can you do so? Play the quiz and assess yourself.

• 1.

### Which of these non-covalent bonds in biological systems is usually the weakest?

• A.

Van der waals interactions

• B.

Ion-dipole interaction

• C.

Hydrogen bond

• D.

Hydrophobic interaction

A. Van der waals interactions
• 2.

### Which of the following is TRUE?

• A.

A buffer cannot be destroyed by adding too much strong base. it can only be destroyed by too much strong acid.

• B.

An effective buffer has a [base]/[acid] ratio the range of 10 -100

• C.

An effective buffer has very small absolut concentrations of acid and conjugate base

• D.

A buffer is most resistant to PH  change when [acid] = [conjugate base]

D. A buffer is most resistant to PH  change when [acid] = [conjugate base]
• 3.

• A.

6.42

• B.

6.40

• C.

6.43

• D.

Option 4

B. 6.40
• 4.

### Which of the following identifies a good buffer?

• A.

Small amount of both a weak acid and its conjugate base

• B.

Significant amount of both a weak acid and its conjugate base

• C.

Significant amounts of both a weak acid and strong acid

• D.

Significant amounts of both a strong acid and strong base

D. Significant amounts of both a strong acid and strong base
• 5.

### Which of the following would be the strongest acid?

• A.

Succinic acid, a diprotic acid with pk=4.21 and 5.64

• B.

Acetic acid pk=4.76

• C.

Formic acid pk=3.75

• D.

Ammonium ion pk=9.25

C. Formic acid pk=3.75
• 6.

### Which of the following statements is FALSE?

• A.

The hydrogen end of water molecules is attracted toward Cl- ions

• B.

Hydration is a special of solvation in which the solvent is water

• C.

Dipole-induced dipole interactions are referred to as london-dispersion forces

• D.

The oxygen end of water molecules is attracted toward ca2+ ions

D. The oxygen end of water molecules is attracted toward ca2+ ions
• 7.

### In a water molecule, hydrogens are partially ___, oxygens are partially  ____

• A.

Negative ; positive

• B.

Positive ;  positive

• C.

Positive ; negative

• D.

Negative ; negative

C. Positive ; negative
• 8.

• A.

PH =4.6

• B.

PH =3.7

• C.

PH =3

• D.

PH =11

C. PH =3
• 9.

### Calculate the pH of the aqueous solutions 1*10^-4M NaOH

• A.

PH=6,7

• B.

PH=11

• C.

PH=10

• D.

PH=8

C. PH=10
Explanation
The correct answer is pH=10. A 1*10^-4M NaOH solution is a weak base. When NaOH dissolves in water, it dissociates into Na+ and OH- ions. The OH- ions combine with water molecules to form hydroxide ions (OH-) and the concentration of OH- ions increases. Since OH- ions are responsible for the basicity of a solution, an increase in their concentration results in a higher pH value. A pH of 10 indicates that the solution is slightly basic.

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• 10.

### All of the above examples on buffer in vitro except?

• A.

TRIS

• B.

Zwitterions

• C.

Phosphate buffer system

• D.

Carbonic acid

D. Carbonic acid
Explanation
The given options are examples of buffers used in vitro. TRIS, Zwitterions, and phosphate buffer system are all commonly used in laboratory settings to maintain a stable pH. However, carbonic acid is not typically used as a buffer in vitro. It is a weak acid that exists in equilibrium with bicarbonate ions in the body, playing a role in maintaining blood pH. Therefore, the correct answer is carbonic acid.

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• 11.

• A.

0.76

• B.

0.63

• C.

1

• D.

0.93

B. 0.63
• 12.

### Using the Henderson-Hasselbalch equation calculate PH when 0.1 mol of NaOH is added pka=4.76 , 0.1 mol of acetic acid react

• A.

3.81

• B.

4.39

• C.

4,76

• D.

5.71

A. 3.81
Explanation
The Henderson-Hasselbalch equation is used to calculate the pH of a solution containing a weak acid and its conjugate base. In this case, acetic acid (CH3COOH) is a weak acid and NaOH is a strong base. When NaOH is added, it reacts with acetic acid to form water and sodium acetate (CH3COONa). The reaction consumes the acetic acid, causing a decrease in the concentration of the weak acid. As a result, the pH of the solution decreases. The pKa value of 4.76 indicates that the solution is slightly acidic. Among the given options, the closest pH value to an acidic solution is 3.81, which is the correct answer.

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• 13.

### Using the Henderson-Hasselbalch equation calculate PH when 0.3 mol of NaOH is added pka=4.76 , 0.3 mol of acetic acid react

• A.

3.81

• B.

4,76

• C.

4.39

• D.

5.71

C. 4.39
Explanation
The Henderson-Hasselbalch equation is used to calculate the pH of a solution containing a weak acid and its conjugate base. In this case, acetic acid (CH3COOH) is the weak acid and its conjugate base is acetate (CH3COO-). When NaOH is added, it reacts with acetic acid to form water and sodium acetate. The equation to calculate pH using the Henderson-Hasselbalch equation is pH = pKa + log([A-]/[HA]), where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid. Given that 0.3 mol of acetic acid reacts, the concentration of acetic acid is reduced by 0.3 mol, while the concentration of acetate is increased by 0.3 mol. Plugging these values into the equation, we get pH = 4.76 + log(0.3/0.3) = 4.76 + log(1) = 4.76. Therefore, the correct answer is 4.76.

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• 14.

### Using the Henderson-Hasselbalch equation, calculate PH when 0.5 mol of NaOH is added pka=4.76 , 0.5 mol of acetic acid react

• A.

3.81

• B.

4.39

• C.

4.76

• D.

5.13

C. 4.76
Explanation
The Henderson-Hasselbalch equation is used to calculate the pH of a solution containing a weak acid and its conjugate base. In this case, acetic acid (CH3COOH) is a weak acid and its conjugate base is acetate (CH3COO-). When NaOH is added, it reacts with acetic acid to form water and sodium acetate. Since NaOH is a strong base, it completely reacts with acetic acid, resulting in the conversion of 0.5 mol of acetic acid to 0.5 mol of acetate. The pKa value given (4.76) is the logarithmic acid dissociation constant of acetic acid. By using the Henderson-Hasselbalch equation, the pH can be calculated as the pKa value, which is 4.76.

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• 15.

### Using the henderson-hasselbalch equation calculate PH when 0.7 mol of NaOH is added pka=4.76 , 0.7 mol of acetic acid react

• A.

3.81

• B.

4.39

• C.

5.13

• D.

6.15

C. 5.13
Explanation
The Henderson-Hasselbalch equation is used to calculate the pH of a solution when a weak acid and its conjugate base are present. In this case, acetic acid (a weak acid) reacts with sodium hydroxide (a strong base) to form sodium acetate and water. The pKa value of acetic acid is given as 4.76. By using the Henderson-Hasselbalch equation, the pH can be calculated using the equation pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base (sodium acetate) and [HA] is the concentration of the weak acid (acetic acid). The equation can be rearranged to calculate the concentration of the conjugate base, [A-]. Since 0.7 mol of acetic acid reacts, the concentration of the conjugate base is also 0.7 mol. Plugging these values into the equation, the pH is calculated to be 5.13.

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