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Hydroxide he would need in order to create 55.59 grams of Copper I phosphate. ( He is reacting Copper I Hydroxide with Phosphoric Acid)
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## Henderson Hasselbalch equation:

Henderson–Hasselbalch equation relates the pH of a solution to the acid dissociation constant, Ka, and the concentrations of the species in solution.

Henderson Hasselbalch equation in buffer calculations, is used to determine the pH of a buffer solution.
The weak acids and weak bases ionize slightly, using this fact we can estimate the pH of the buffer solutions using initial concentrations.

## Henderson Hasselbalch equation Sample Questions:

Question 1: If you know the pKa of an acid, you can use the Henderson Hasselbalch equation to determine the __ of a __

a) pKb, Base                            b) pKa, Acid
c) pH, Solution                         d) pKa, Ion.

Explanation: Knowing the pKa of an acid, Henderson Hasselbalch equation can be used to calculate the pH of a solution.

Question 2: A buffer solution comprises which of the following?

a) A weak acid in solution                             b) A strong acid in solution
c) A weak base in solution                            d) A weak acid and its conjugate base in solution

Answer: Option d) A weak acid and its conjugate base in solution

Explanation: A buffer solution consists of a weak acid and its conjugate base (or a weak base and its conjugate acid). It cannot contain the weak acid or weak base in isolation.

Question 3: pKa in Henderson Hasselbalch  represents-

a) Density                                      b) Pressure
c) Acidic strength                           d) Basic strength

Explanation: pKa indicates the dissociation constant and how easily a proton can be extracted from an acid. Therefore pKa represents acidic strength.

Question 4: A solution of acetic acid (pKa = 4.75) has a pH of 6.75. The ratio of acid to conjugate base is…….

a) 100 CH3COOH to 1 CH3COO–                            b) 1 CH3COOH to 100 CH3COO–
c) 100 CH3COO– to 1 CH3COOH                            d) 1 CH3COO– to 100 CH3COOH

Answer: Option b) 1 CH3COOH to 100 CH3COO–

Explanation: Applying Henderson Hasselbalch equation-

Question 5: To produce a buffer with pH of 5.75, You have a solution with 30.0g of acetic acid (pKa=4.75). How many moles of sodium acetate must be added to achieve the desired pH?

a) 5.0mol                            b) 3.0mol
c) 0.5mol                            d) 1.3mol

Explanation: Using Henderson Hasselbalch equation-

Assuming these ions occupy same volume

We know 30 g of acetic acid is equal to 0.5 mol, putting in values-

=5 mol

Question 6: Solution with small values of pKa are-

a) Strong acids.                             b) Weak acids.
c) Strong bases.                            d) Buffers

Explanation: The smaller the pKa values of an acid, the more easily it donates its proton. Hence solutions with small pKa values are strong acids.

Question 7: The Henderson–Hasselbalch equation

a) Allows one to determine the molecular weight of a weak acid from its pH alone.
b) Is equally useful with solutions of acetic and hydrochloric acid.
c) Employs the same value for pK for all weak acids.
d) Relates the pH of a solution to the pK and the concentrations of weak acid and conjugate base.

Answer: Option d)  Relates the pH of a solution to the pK and the concentrations of weak acid and conjugate base

Explanation: Henderson–Hasselbalch equation relates the pH of a solution to the pKa and the concentrations of weak acid and conjugate base

Question 8: Which of the following statements about buffers is true?

a) The pH of a buffered solution remains constant no matter how much acid or base is added to the solution
b) The strongest buffers are those composed of strong acids and strong bases
c) The maximal buffering capacity occurs when pH = pK
d) The pH of blood is not maintained by a buffering system

Answer: Option c) The maximal buffering capacity occurs when pH = pK

Explanation: Buffer capacity of a buffer is maximum when the concentration of the weak acid and its salt or weak base and its salt are equal i.e., when pH=pKa or pOH=pKb.

Question 9: Which of the following is true regarding the Henderson-Hasselbalch equation?

a) The pH of the solution is always greater than the pKa of the solution.
b) As the ratio of conjugate base to acid increases, the pH increases.
c) The hydrogen ion concentration can never equal the acid dissociation constant.

Answer: Option b) As the ratio of conjugate base to acid increases, the pH increases

Explanation: Increasing the ratio of [A-] to [HA] will increase the logarithm, and the pH of the solution. This is because conjugate base is more than acid, making solution more alkaline and increased pH.

### When can you use the Henderson Hasselbalch equation?

We can use Henderson Hasselbalch equation in buffer calculations, to determine the pH of a buffer solution.

pH=pKa+log10([A-]/HA)

The weak acids and weak bases ionize slightly, using this fact we can estimate the pH of the buffer solutions using initial concentrations.

### Henderson-Hasselbalch equation is used to calculate?

The Henderson–Hasselbalch equation is used to calculate the pH of a buffer solution in acid base reactions.

pH=pKa+log10([A-]/HA)

pKa represents acid dissociation constant.

[A-]and [HA] represents concentrations of conjugate base and the starting acid respectively.

### Which statements regarding the Henderson-Hasselbalch equation are true?

1) If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be calculated.

2) If pH = pK a for an acid, the amount of conjugate base is equal to the amount of acid in solution.

3) If pH >> pK a for an acid, the acid will be mostly ionized.

4) If pH << pK a for an acid, the acid will be mostly ionized.

Statement 1,2,3 are true, statement 4 is false

The Henderson-Hasselbalch equation is:

pH=pKa+log10([A-]/HA)

If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. TRUE

pH=pKa+log10([A-]/HA)

If you know pHand pKa

10pH-pKa=([A-]/HA)

At pH = pKa for an acid, [conjugate base] = [acid] in solution. TRUE

pH=pKa+log10([A-]/HA)

0 = log10([A-]/HA)

100=[A-]/HA

1 = [A-]/HA

[conjugate base] = [acid] in solution.

At pH >> pKa for an acid, the acid will be mostly ionized. TRUE

pH=pKa+log10([A-]/HA)

If pH >> pKa, 10pH-pKa will be >> 1, which means more [A⁻] than [HA]

At pH << pKa for an acid, the acid will be mostly ionized. FALSE

pH=pKa+log10([A-]/HA]

If pH << pKa, 10pH-pKa will be << 1, which means more [HA] than [A⁻]

### How is the Henderson-Hasselbalch equation related to solubility?

Henderson-Hasselbach equation gives the relation between pH, pKa, and the concentrations of an acid and its salt-

pH=pKa+log10([A-]/HA)

When pH decreases, the concentration of the acid increases and that of the salt decreases. The undissociated form is much less soluble than its salt. Change in solubility brought about by change in pH can be predicted by the pHp equation.

pHp=pKa+logs-s0/s0(for weak acid)

pHp=pKw-pKb+logs0/s-s0(for weak base)

pHp is the pH value below which an acid or above which a base will start to precipitate.