People often take milk of magnesia to reduce the discomfort associated with acid stomach or heartburn. The recommended dose is 1 teaspoon, which contains 4.00 * 102 mg of Mg(OH)2. What volume of an HCl solution with a pH of 1.3 can be neutralized by one dose of milk of magnesia? If the stomach contains 2.00 * 102 mL of pH 1.3 solution, is all the acid neutralized? If not, what fraction is neutralized?

Video transcript

Hello everyone today, we have been given the following scenario and asked to analyze it. So it says laboratory preparation of carbon dioxide gas is often carried out by reacting limestone with hydrochloric acid. The limestone dissolves completely into the acid solution and produces C. A. C. L. To water. And carbon dioxide were then asked to determine the volume. And middle leaders of the ph of 1.1 hydrochloric acid solution that could dissolve a 1.15 grand piece of limestone. Could the same piece of limestone neutralize a 350 sample of hcl solution with a ph of 1.1, if not what fraction can neutralize. And so the first thing we want to do is we want to denote, how can we find the hydro knee um concentration. So we can remember or recall that the hydrogen concentration is equal to 10 raised to the negative ph And so when we do this we have 10 raised to the negative ph or in other words negative 1.1. And this gives us a hydrogen concentration of 0.794 moller. And since hcl is a very strong acid, our hydrogen concentration is equal to how much a cl hcl that we have. And so the reaction equation in the situation here is CAC 03, a solid. And we're adding two equivalents of hcl in the acquis form. And this is going to give us a C a C. L. To acquis plus water in a liquid form of course plus C. +02 gas. And so now we have to determine the volume of hcl total. And so before we do that, we wanted to note that the molar mass of C a c 03 is 100.9 g per moller. And so to find the volume of Rh ceo. So to find the volume of hcl, we're going to take how many grams of C a c 03 that we have, which is 1.15 g. And then we're going to use the molar mass conversion factor. We're going to say one mole of this. C a C 03 is equal to 100. g. And then we're going to multiply by the multiple ratio. And we're going to say that for every one mole of c a c 03, we have two moles of H C L. Then we are going to say that for every 0.0794 moles of Hcl, we have one leader of hcl And then of course we have to convert to middle leaders. And by doing that, we say that for every one middle leader, we have 10 to the negative third leaders of Hcl. And once our units cancel out, we're going to end up with 289 mL as our volume. The other solution I stated in the question is 350 middle leaders. And so therefore it cannot be completely neutralized. Right? So therefore the fraction. So the fraction That is neutralized can be found by simply taking the 289 ml and dividing that by the 350 ml sample to give us 0.8-6 ml. And that is our final answer. I hope this helped until next time.