(1b) For a particular experiment, Candy Chemist, our favorite acid-base buffer s

Question

(1b) For a particular experiment, Candy Chemist, our favorite acid-base buffer savant, must prepare a buffer with a pH = 8.50. In the laboratory, she finds 1.00M acetic acid 1.76x10-5) and 1.00 M ammonia (K = 1.76x10-5). Which reagent should be used to prepare the buffer? 1.DD (Ka = In addition to the above question, your assignment is to calculate the mole ratio of acid to conjugate base (or base to conjugate acid), which can be used to determine the best reagent to prepare the buffer. Is the buffer better for additions of acid or base? Explain.

Explanation / Answer

Ans. Part 1: The buffering capacity of a ‘weak acid – conjugate base’ OR ‘weak base- conjugate acid’ buffer is maximum when pKa of the weak acid is closes to the specified pH of buffer.

# Given, Ka of acetic acid = 1.76 x 10-5

Now, pKa of acetic acid = -log Ka = -log (1.75 x 10-5) = 4.75

# Given, Kb of ammonia = 1.76 x 10-5

NH4+ is the conjugate acid of NH3 base.

Now, Ka of NH4+ = 10-14 / Kb = 10-14 / (1.76 x 10-5) = 5.6818 x 10-10

Now, pKa = -log (5.6818 x 10-10) = 9.25

# Since pKa of NH4+ ( or, base ammonia (9.25) is closest to that of the desired buffer pH 8.50, NH3 is the best choice for preparing the buffer.

Hence, ammonia should be used to prepare the buffer.

# Part 2: We have-               Base = NH3    ; conjugate acid = NH4+      ; pH = 8.50

Now,   pOH = 14.00 – pH = 14.00 – 8.50 = 5.50

            pKb = -log (1.75 x 10-5) = 4.75

# Using Henderson-Hasselbalch equation for weak acid, pOH = pKb + log ([BH+] / [B]);        Where                        B = Base , BH+ = Conjugate acid

Putting the values in above equation –

            5.50 = 4.75 + log ([BH+] / [B])

            Or, log ([BH+] / [B]) = 5.50 – 4.75 = 0.75

            Or, [BH+] / [B] = antilog 0.75

            Or, [BH+] / [B] = 5.62

            Hence, [B] / [BH+] = 1 / 5.62 = 0.18

Therefore, [B] / [BH+] = 0.18

# Part 3: Since the relative amount of base is far less than (around 5 times lesser) than its conjugate base, the conjugate acid can donate more protons to basic species being added to it to maximize buffering capacity. Buffering capacity is maximum when concentration of conjugate base and base is equal to each other. Therefore, the buffer is best suited for addition of base.

# If an acidic species is added to the buffer, the [BH+] increase and [B] would decrease. The greater is the difference in [BH+] and [B] in the buffer, lower would be the buffering capacity. So, addition of an acid would decrease the buffering capacity of the buffer, Hence, addition of acid is NOT preferred for this buffer.

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