The daily output of stomach acid (gastric juice) is 1000 mL to 2000 mL. Prior to a meal, stomach acid (HCl) typically has a pH of 1.42.
c. The antacid milk of magnesia contains 400. mg of Mg(OH)₂ per teaspoon. Write the neutralization equation, and calculate the number of milliliters of stomach acid neutralized by 1 tablespoon of milk of magnesia (1tablespoon = 3teaspoons).

Question

The daily output of stomach acid (gastric juice) is 1000 mL to 2000 mL. Prior to a meal, stomach acid (HCl) typically has a pH of 1.42.

c. The antacid milk of magnesia contains 400. mg of Mg(OH)₂ per teaspoon. Write the neutralization equation, and calculate the number of milliliters of stomach acid neutralized by 1 tablespoon of milk of magnesia (1tablespoon = 3teaspoons).

Accepted Answer

Welcome back everyone. The volume of an hydrochloric acid waste is 10,000 mL. It has a ph of two point oh and a small bottle contains one g of calcium hydroxide. Show the neutralization reaction between hydrochloric acid and calcium hydroxide and determine how many milliliters in one significant figure of hydrochloric acid waste can be neutralized by two small bottles of calcium hydroxide. We're going to begin by recognizing that hydrochloric acid is one of our memorized strong acids from our textbook and calcium hydroxide is one of our memorized strong bases from our textbooks. When we have the reaction between a strong acid and a strong base, this will be known as a neutralization reaction. So showing that we're going to have hydrochloric acid react with calcium hydroxide. Since these are both strong reactants, as far as being a strong acid and a strong base, they will fully dissociate into ions. And because this is a neutralization reaction, we have a full reaction arrow to our products which will be a salt and then water. Recall that the salt, the formula for the salt will be determined by the combination of the ion from our base in this case, that would be the dissociated calcium two plus ion, which will combine with the anion from our acid being our chloride one minus anion. Recall that because calcium forms a plus two charge, because calcium is found in group two a of the periodic table that plus two charge will become a subscript for the anion. So we would have C AC L two as the formula for the salt. Now, we need to ensure that we have a balanced equation. So we're going to compare the atoms on both sides of the reaction so that they can match up, we have on both sides, hydrogen chlorine, calcium and oxygen. And starting with the reactant side, we have a total of three atoms of hydrogen, one atom of chlorine, one atom of calcium and two atoms of oxygen going to our product side. We observe that we have a total of two atoms of hydrogen, two atoms of chlorine, one atom of calcium and one atom of oxygen. Recall that to make things simple when balancing, you start by balancing non hydrogen and non oxygen atoms first. So since we can observe that our chlorine atoms are imbalanced where we have two on the product side and just one atom on the reactant side, we're going to fix that by placing a coefficient of two in front of our reactant hydrochloric acid, which is now going to give us two atoms of chlorine on both sides of the reaction. And with that coefficient introduced, we have now changed our number of hydrogen on the reactant side to a total of four atoms. So now with that change, we're going to then focus on balancing hydrogen and oxygen. Now because we only have two atoms of hydrogen on the product side, we're going to place a coefficient of two in front of our products. Water that's now going to balance out our hydrogen so that we have four atoms on both sides of the equation. And then our oxygen will also be balanced with two atoms on both sides of the equation. So now we would have a fully balanced reaction. And our next step because we need to find the volume in milliliters of our strong acid H C L that is neutralized by the strong base. And we are given a P H of the solution of H C L East. We can find the concentration of protons that are equivalent to that P H. And so recall our P H formula where P H is calculated by taking the negative log of our concentration of protons H plus, we can isolate this formula to solve for the concentration of H plus plugging in are given ph by saying that the concentration of H plus is found by taking 10 to the negative P H. In this case, we would have 10 to the negative two point oh. So simplifying in our calculators that would yield a concentration of protons of 0.1 molar. And as we stated before, hydrochloric acid is a strong acid and therefore fully dissociates into its constituent ions being a proton H plus and the chloride one minus anion. And so therefore, the concentration of hydrochloric acid is going to equal the concentration of protons H plus and the concentration of chloride. And that's because we have that 1 to 1 ratio of the fully dissociated ions with their parent acid H C L. So because we were able to use the given P H to find the concentration of protons, then those concentrations of each of our species H C L H plus and C L minus is again 0.1 molar. And recall that molar capital m is interpreted in terms of moles of solute per liter of solution. So we can say 0.1 moles per liter as an alternative. So the next step is now that to get to our final part of our answer is we're going to finally solve for that volume of hydrochloric acid that can be neutralized by the strong base. And so we're going to need to take the inverse of the concentration of our strong acid H C L. So that we can interpret its concentration as 0.1 moles in the denominator. And this is per one liter of hydrochloric acid solution. We want the, the concentration to be set as an inverse because we want our final units to be milliliters. So we need leaders to be in the numerator. And we're going to begin by using dimensional analysis and multiply to go from one liter of hydrochloric acid solution in the denominator to then one mL of hydrochloric acid solution in the numerator. This will then allow us to cancel out liters of hydrochloric acid solution. And now we're going to focus on utilizing the coefficients from our balanced equation above where we referenced that for every two moles of hydrochloric acid reacted, we have two or one mole of calcium hydroxide reacted as well. And so we can multiply then to go from moles of H C L in the denominator, two moles of H C L in the numerator. And we want to get to moles of our strong base calcium hydroxide in the denominator. And so referring to our ratio above, we said for every two moles of H C L reacted, we have one mole of calcium hydroxide also reacted. So canceling out now moles of H C L, we're then going to multiply to incorporate the molar mass of calcium hydroxide. So we would want moles of calcium hydroxide in the new and then grams of calcium hydroxide in the denominator recall that utilizing our periodic table, we've got a mass of 79 0.9 g of calcium hydroxide per mole. So canceling out grams or sorry, canceling out moles of calcium hydroxide. We're then going to multiply to incorporate the information from the prompt where we are told that for the total of two bottles of the strong base calcium hydroxide, we're going to multiply to include the conversion factor that for every bottle in the denominator. So per bottle in the denominator of calcium hydroxide, there is a mass of one g of calcium hydroxide and so canceling out. Now bottles, we are left with also grams of calcium hydroxide to cancel out. And our final unit is going to be milliliters of the hydrochloric acid solution. And so in our calculators, this is going to simplify so that we find a volume of 5399 millilitres. I'm sorry, just to make a correction for our first multiplier conversion factor going from liters of H C L solution to milliliters of H C L solution. We need to recall that the conversion is for every liter we have 10 to the third milliliters. And so that will allow us in our calculators to get mL because we need to round our answer to one significant figure. We're going to just write it out as milliliters of H C L solution that can be neutralized. And so our final answer is 5000 mL. I hope that this made sense and let us know if you have any questions

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Key Concepts

Neutralization Reaction

Molarity and Concentration

Volume Calculation