Question

17

You may want to reference (Page) section 16.2 while completing this problem.

Use the Henderson-Hasselbalch equation to calculate the pH of each solution:

Part A

a solution that is 0.170 M in HC2H3O2 and 0.125 M in KC2H3O2

Express your answer using two decimal places.

pH=

Part B

a solution that is 0.205 M in CH3NH2 and 0.135 M in CH3NH3Br

Express your answer using two decimal places.

pH=

Answer 1

Part A) Given, a buffer solution that is 0.170 M in HC₂H₃O₂ and 0.125 M in KC₂H₃O₂

We know the K_a value for acetic acid = 1.8 x 10^-5

Thus, pK_a = -logK_a

pK_a = -log(1.8 x 10^-5)

pK_a = 4.7447

Now, the equilibrium representation of this solution is,

HC₂H₃O₂(aq) + H₂O(l) C₂H₃O₂^-(aq) + H₃O⁺(aq)

We know, the Henderson-Hasselbalch equation,

pH = pK_a + log [C.base / Acid]

Substituting the known values,

pH = 4.7447 + log [0.125 / 0.170]

pH = 4.7447 - 0.1335

pH = 4.61

Part B)

Given, a buffer solution that is 0.205 M in CH₃NH₂ and 0.135 M in CH₃NH₃Br

We know, the K_b value for CH₃NH₂ = 4.4 x 10^-4

Thus, pK_b = -logK_b

pK_b = -log(4.4 x 10^-4)

pK_b = 3.36

Now, the equilibrium representation of this solution is,

CH₃NH₂(aq) + H₂O(l) CH₃NH₃⁺(aq) + OH^-(aq)

We know, the Henderson-Hasselbalch equation,

pOH = pK_b + log [C.Acid / Base]

Substituting the known values,

pOH = 3.36 + log [0.135 / 0.205]

pOH = 3.36 - 0.1814

pOH = 3.1751

Now, We know,

pH + pOH = 14

pH + 3.1751 = 14

pH = 10.82