Chemistry, asked by activem299, 8 months ago

Which buffer system is the best choice to create a buffer with pH pH = 7.20? (A) HC, H, O, KC, H, O, (B) HCIO, / KCIO, (C) HCIO / KCIO (D) NH / NH, Cl Answer OA B D Submit​

Answers

Answered by sarbjitsingh75890
14

Answer:

Part A: Which of the following buffer systems would be the best choice to create a buffer of pH = 7.20?

(a) CH3COOH & CH3COOK 

(b) HClO2 & KClO2 

(c) NH3 & NH4Cl 

(d) HClO & KClO

Part B: For the best system, calculate the ratio of masses of the buffer components required to make the buffer.

Solution to A:

1) Examine the pKa of each acid:

(a) Ka = 1.77 x 10¯5; pKa = 4.752 

(b) Ka = 1.1 x 10¯2; pKa = 1.96 

(c) Ka = 5.65 x 10¯10; pKa = 9.248 

(d) Ka = 2.9 x 10¯8; pKa = 7.54

2) Write the Henderson-Hasselbalch equation:

pH = pKa + log [base / acid]

3) Set the base/acid ratio to 1/1 and the result is:

pH = pKa

4) Look for the pKa value nearest to a pH of 7.20 when the base/acid ratio is 1/1:

the best choice for a pH = 7.20 buffer is answer choice d, the HClO & KClO buffer

5) Why?

The buffer with a pKa closest to 7.20 would be the best choice because it has the least adjusting to do in the base/acid ratio in order to reach a buffer of pH = 7.20. I will return to this below.

Solution to B:

1) Let's write the Henderson-Hasselbalch equation again:

pH = pKa + log [base / acid]

2) We know the desired pH of the buffer, so let's put it in:

7.20 = pKa + log [base / acid]

3) Let's get the pKa of HClO:

I will use a pKa equal to 7.5376 (I'll keep a couple extra digits for the moment.)

4) Put it in place:

7.20 = 7.5376 + log [base / acid]

log [base / acid] = −0.3376

[base / acid] = 0.45962 (still keeping some extra digits)

5) The above ratio is a ratio of moles, but we want a ratio of weight. What we will do is use the molar masses of KClO and HClO and change the mole ratio to a weight ratio.

numerator (KClO) = (0.45962 mol) (90.55 g/mol) = 41.6186 g 

denominator (HClO) = (1 mol) (52.4599 g/mol) = 52.46 g

The weight ratio is 41.6186 g / 52.46 g = 0.79 (to two sig figs)

Explanation:

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