The pH in your stomach varies considerably depending on whether or not food is b
ID: 565509 • Letter: T
Question
The pH in your stomach varies considerably depending on whether or not food is being actively digested. During digestion, stomach pH typically ranges from 1-2, while resting pH generally ranges from 3-4. Antacids are a class of medicines which help neutralize stomach pH, particularly for those who might experience acid reflux. A common chemical component of antacids is aluminium hydroxide – Al(OH)3, a single molecule of which can neutralize 3 free hydrogens. Assume that each dose of your antacid medicine contains 300mg of aluminium hydroxide (MW = 78 g/mol) and that the volume of your stomach is 0.90L.
Part A) Let’s say that after you’ve eaten you know that the pH in your stomach is 1.50 and that you will experience acid reflux until the pH in your stomach is at least 4.0. How many doses of antacid would you need to take in order to raise your pH to or above 4.0 (assume you can only take full doses – no partial doses)?
Part B) Assuming that the antacid has a clearance rate from the stomach of 1 mM/hr and that stomach acid (H+) is produced at a rate of 0.5 mM/hr, and that you took one more dose of antacid than you needed to get your stomach pH above 4 (add 1 to your previous answer), for how long will the antacid you took keep your stomach pH above 4? Assume steady state clearance of the antacid
Explanation / Answer
Ans. Part A: Given, Volume of stomach (= stomach acid) = 0.90 L
pH of stomach = 1.50
Required pH after taking antacid = 4.0 or above
# Initial [H3O+] in stomach = 10-pH = 10-1.5 = 0.031623 M
Initial moles of H3O+ = Initial [H3O+] x Volume of stomach acid in liters
= 0.031623 M x 0.90 L
= 0.0284607 mol
# Final [H3O+] in stomach = 10-pH = 10-4.0 = 0.0001 M
Final moles of H3O+ = Final [H3O+] x Volume of stomach acid in liters
= 0.0001 M x 0.90 L
= 0.00009 mol
# Moles of H3O+ to be neutralized = Initial moles of H3O+ - Final moles of H3O+
= 0.0284607 mol - 0.00009 mol
= 0.0283707 mol
# 1 mol Al(OH)3 neutralizes 3 mol H3O+. So,
Required moles of Al(OH)3 = (1/3) x moles of H3O+ to be neutralized
= (1/3) x 0.0283707 mol
= 0.0094569 mol
Required mass of Al(OH)3 = Required moles x molar mass
= 0.0094569 mol x (78 g/ mol)
= 0.7376382 g
= 737.6382 mg
# Given, Al(OH)3 content of antacid = 300 mg Al(OH)3 per tablet or dose
Required dose of antacid = Required mass of Al(OH)3 / (Al(OH)3 content of antacid)
= 737.6382 mg / (300 mg/ tablet)
= 2.458794 tablet
Since pH shall be 4.0 or above, the number of antacid acid doses can’t be less than 2.45, i.e. it can’t be rounded off to 2 tablets or dose.
Since the dose must be in whole number-
The required number of antacid dose = 3
# Part B. Given-
Total number of antacid dose taken = 3 + 1 = 4
Number of extra dose of antacid taken = Dose taken – dose actually required
= (4 - 2.458794) dose
= 1.541206 dose
Spontaneous clearance of antacid Al(OH3) = 1 mM/hr
Rate of [H+] production by stomach = 0.5 mM/ hr
# 1 mol H+ neutralizes 1 mol OH-. Since Al(OH)3 has 3 moles of OH-, 1 mol gastric H+ neutralizes (1/3 mol = 0.333 mol) antacid Al(OH)3.
So,
Moles of antacid neutralized by gastric H+ per hour = (1/3) x rate of H+ formation
= (1/3) x (0.5 mM/hr)
= 0.1667 mM/hr
# Total antacid removal rate =
Spontaneous rate of removal + Rate of neutralization of gastric H+
= 1.0 mM/hr + 0.1667 mM/hr
= 1.1667 mM/hr
# Amount of excess antacid above pH 4 taken =
1.541206 dose x antacid content per dose
= 1.541206 dose x (300 mg/ dose)
= 462.3618 mg
= 0.4623618 g
= 0.4623618 g / (78 g/ mol)
= 0.0059277 mol
= 5.9277 mmol
Given, volume of stomach = 0.90 L
So, molarity of excess antacid = mmole of excess antacid / Volume in liters
= 5.9277 mmol/ 0.90 L
= 6.586 mmol/ L
= 6.586 mM
# Required time to bring pH to 4.0 =
Excess of antacid taken / Total rate of antacid removal
= 6.586 mM / (1.1667 mM/hr)
= 5.645 hr
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