You are working in a candy factory that makes both cherry and lemon candy. One a

Question

You are working in a candy factory that makes both cherry and lemon candy. One afternoon after lunch, you get sleepy and accidentally put the ingredients for cherry candy into the lemon candy vat. You remember from your bioengineering class that sour taste receptors detect hydrogen ion concentration to signal a sour taste, and that sweet taste receptors detect organic substances. You decide to fix your problem by simply adjusting the pH of the lemon vat. The pH of an uncontaminated vat of liquid lemon candy is 2.85. The pH of the contaminated lemon candy liquid is 3.4, and the vat contains 50 gallons of liquid. Assume that the contaminated lemon candy liquid behaves as a strong acid. (a) You first think about adjusting the pH using 0.25 M hydrochloric acid. What volume of hydrochloric acid should you add in order to correct the mistake? (b) You then decide that HCI probably tastes bad and probably shouldn't be ingested, so you change your acid to 1.0 M ascorbic acid (vitamin C), which has a pKa of 4.17. (For this problem, assume only one dissociation). How much ascorbic acid should you add in order to correct your mistake? (c) How would the calculated volume change if the ascorbic acid were 2.0 M: One of the main functions of saliva is to buffer against acid from food and plaque, which contributes significantly to the formation of cavities. White there are several buffers in the saliva, carbonic acid (H_2CO_3) has the highest concentration and has the greatest effect on pH.

Explanation / Answer

Question 5.36 a

pH of uncontaminated lemon candy liquid is 2.85.

pH is log (1 / [H+] ). Hence, in uncontaminated lemon candy liquid, log (1 / [H+] ) = 2.85

1 / [H+] = antilog 2.85 = 7.079 * 102

[H+] = 1 / (7.079 * 102) = 1.413 * 10-3 M

Hence, in uncontaminated lemon candy liquid, [H+] = 1.413 * 10-3 M         ..............(1)

pH of contaminated lemon candy liquid is 3.4

So, in contaminated lemon candy liquid, log (1 / [H+] ) = 3.4

1 / [H+] = antilog3.4 = 2.512 * 103

[H+] = 1 / (2.512 * 103) = 3.981 * 10-4 M

Hence, in contaminated lemon candy liquid, [H+] = 3.981 * 10-4 M         ..............(2)

Given, concentration of hydrochloric acid = 0.25 M

Certain volume of 0.25 M hydrochloric acid is to be added to contaminated lemon candy liquid so that [H+] in final solution is equal to that in uncontaminated lemon candy liquid i.e 1.413 * 10-3 M (result 1).

Molarity with respect to H+ ions in contaminated lemon candy liquid, M1 = 3.981 * 10-4 M  (result 2)

Volume of contaminated lemon candy liquid, V1 = 50 gallons = 189.2705 L     (since 1 gallon = 3.78541 L)

Molarity of hydrochloric acid, M2 = 0.25 M

Let volume of 0.25 M hydrochloric acid required be V2

Molarity of final solution required with respect to H+ ions is MFINAL = 1.413 * 10-3 M

(Aim is that MFINAL should be same as concentration of H+ ions in uncontaminated lemon candy liquid)

Total volume of final solution, VFINAL = V1 + V2 = 189.2705 + V2    ......L

(MV)CONTAMINATED LEMON CANDY LIQUID + (MV)HCl = (MV)FINAL

M1V1 + M2V2 = MFINAL VFINAL

(3.981 * 10-4 * 189.2705) + (0.25 * V2) = (1.413 * 10-3) * (189.2705 + V2 )

(0.25 * V2) - (1.413 * 10-3 * V2) =  (0.26744) - (0.07535)

0.2486 * V2 = 0.1921

Hence, V2 = 0.7727 L = 772.7 mL           ( 1L = 1000 mL)

Thus, volume of HCl to be added to correct the mistake is, 0.7727 L or 772.7 mL

__________________________________________________________________________________

Question 5.36 b

Given pKa of ascorbic acid is 4.17.   pKa = log (1/Ka ) where Ka is acid dissociation constant.

So, log (1/Ka ) = 4.17    or (1/Ka ) = antilog 4.17 = 1.479 * 104

Ka = 1 / (1.479 * 104 ) = 6.761 * 10-5

Hence, acid dissociation constant of ascorbic acid is Ka = 6.761 * 10-5

Given, concentration of ascorbic acid is, C = 1M

[H+] in ascorbic acid solution can be calculated using the expression, [H+] = (Ka * C)1/2

[H+] = (Ka * C)1/2 = (6.761 * 10-5 * 1)1/2 = 8.22 * 10-3 M

Hence, [H+] in 1M acorbic acid solution is, M3 = 8.22 * 10-3 M

Certain volume of 1M ascorbic acid solution is to be added to contaminated lemon candy liquid so that [H+] in final solution is equal to that in uncontaminated lemon candy liquid i.e 1.413 * 10-3 M (result 1 under answer for question a).

We have, [H+] in 1M acorbic acid solution is, M3 = 8.22 * 10-3 M

Let volume of ascorbic acid solution required be V3

Molarity with respect to H+ ions in contaminated lemon candy liquid, M1 = 3.981 * 10-4 M  (result 2 under answer for question a)

Volume of contaminated lemon candy liquid, V1 = 50 gallons = 189.2705 L     (since 1 gallon = 3.78541 L)

Molarity of final solution required with respect to H+ ions is MFINAL = 1.413 * 10-3 M

(Aim is that MFINAL should be same as concentration of H+ ions in uncontaminated lemon candy liquid)

Total volume of final solution, VFINAL = V1 + V3 = 189.2705 + V3   ......L

(MV)CONTAMINATED LEMON CANDY LIQUID + M3V3 = (MV)FINAL

M1V1 + M3V3 = MFINAL VFINAL

(3.981 * 10-4 * 189.2705) + (8.22 * 10-3 * V3) = (1.413 * 10-3) * (189.2705 + V3)

V3 * {(8.22 * 10-3 ) - ( 1.413 * 10-3 )} =  (0.26744) - (0.07535)

V3 * (6.807 * 10-3 ) = 0.1921

Hence, V3 = 28.221 L

Thus, volume of 1 M ascorbic acid solution to be added to correct the mistake is, 28.221 L

__________________________________________________________________________________

Question 5.36 c

If ascorbic acid were 2M then, [H+] = (Ka * C)1/2 = (6.761 * 10-5 * 2)1/2 = 11.625 * 10-3 M

[H+] in 2M acorbic acid solution is, M4 = 11.625 * 10-3 M

M1V1 + M4V4 = MFINAL VFINAL

(3.981 * 10-4 * 189.2705) + (11.625 * 10-3 * V4) = (1.413 * 10-3) * (189.2705 + V4)

V4 * {(11.625 * 10-3 ) - ( 1.413 * 10-3 )} =  (0.26744) - (0.07535)

V4 * (10.212 * 10-3 ) = 0.1921

Hence, V4 = 18.811L

Thus, volume of 2 M ascorbic acid solution to be added to correct the mistake is, 18.811L

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