A 1.000 L sample of HCl gas at 25°C and 732.0 mm Hg was absorbed completely in an aqueous solution that contained 6.954 g of Na2CO3 and 250.0 g of water. (a) What is the pH of the solution?

Video transcript

hi everyone for this problem. It reads a 0.275 liter sample of hydrogen bromide was dissolved in a solution at 25 degrees Celsius and 730 millimeters of mercury. The solution initially contained 1. g of sodium phosphate and 0. liters of water, calculate the P. H. Of the resulting solution if the P. K. A. Of di hydrogen phosphate is 7.21. So the question that we want to answer here is the P. H. Of the resulting solution. Okay. And let's take a look at what were given in the problem and the problem, we know what the volume is for hydrogen bromide gas. We know what the temperature is, and we know what the pressure is. And we also know what the mass is for our di nitrogen for our di sodium phosphate. Excuse me. And also the volume. Okay, So, and we're also given the P. K. Alright, So in order for us to calculate the P. H. Because we have a gas that we're dealing with, we need to first solve for the number of moles of hydrogen bromide gas and we can do that by using the ideal gas law, which is P. V equals N R. T. Okay, And because we want to solve for the number of moles, we're going to isolate this variable and when we do that, it becomes N. Is equal to P V over R. T. So let's go ahead and solve for the moles of hydrogen bromide. Okay, so we'll go ahead and plug in for pressure. We have 730 millimeters of mercury. We're going to want to convert this millimeters of mercury two atmospheres. So we're going to do is multiply by one atmosphere Over of mercury. That's the conversion that we're going to use. And when we when we use that conversion, the units of mm of mercury are going to cancel. Okay, and then we're going to multiply by volume. And the problem we're told the volume is 0. liters. Okay, and this is all over R times T R is a gas constant that we should have memorized, which is 0. leaders atmosphere over mole kelvin And this is times the temperature. Okay. And for the temperature, we're told that it is 25 degrees Celsius and we need to convert this degrees Celsius to kelvin. So we're going to do is we're going to take this 25 degrees Celsius and add 273.15. Okay, so then we're going to get n or the number of moles is equal to 0.107963. Okay, And this is the moles of hydrogen bromide gas. Okay, so now that we know the number of moles, the reaction that we have for hydrogen bromide gas when it reacts with water is it is going to produce hydro ni um ions plus the an ion, which means hydrogen bromide completely dissociates. So what that means then is the number of moles of hydrogen bromide is going to equal the number of moles of are hydro ni um ions. So this number is the same. So that is 0. 963 moles. Okay, so now that we know the number of moles of hydrogen bromide and hydro me, um we're going to now look at what we have in terms of the dye sodium phosphate. Okay, So in the problem we're told the solution initially contained 1.740 g of di sodium phosphate. So let's go ahead and use that as our starting point, We have 1.740 g. And what we want to do is calculate the number of moles of di sodium phosphate we have. So we need the molar mass here. So we want to go from grams to moles and in one mole We need to use the molar mass here in one mole, there is 141. g. Okay, so the units for grams cancel and we're left with moles. So when we calculate the moles, this equals 0. moles. Okay, so for our di hydrogen, excuse me, are di sodium phosphate. When this dissociates, let's go ahead and write out that. So we have when this dissociates, it associates into the following ions and it dissociates completely. So what that means then is the moles or I'll write it the same way as above. So the moles of di sodium phosphate equals the moles of the conjugate base. And that number is the same. So 0. moles. And we know that everything dissolved in 0.125 liters of water. So we can calculate the mill arat E. Okay, so to calculate the mill arat E, we're going to say that the mill arat E of the conjugate base Is going to equal the moles over leader the moles of over the volume. Okay, so we have 0. moles over the volume, Which is 0.125 l of water. So this gives a mil arat e. 0.098 moller for the conjugate base and Armel arat E. Of hydro ni um Then is going to equal its moles over the volume of water. So 0. moles over 0.125 L of water. So this gives a malaria t. of 0. 704 molar. Okay, so with this now we can create an ice table because the ice table is what's going to help us to solve for P. H. So to start off with our ice table, we need to write out the reaction and the reaction is going to be a reaction of the conjugate base reacting with hydro ni um Okay so what that looks like when we write it is the following. Okay so for our ice table, remember we write out ice on the side I change for I stands for initial concentration. Speech stands for change in concentration and E stands for equilibrium. Okay and so let's go ahead and fill in the table. And remember for Ice tables and for equilibrium expressions we don't include solids or liquids. So this water here we're going to go ahead and ignore it. And I like to draw a line here to separate the products from the reactant. So let's go ahead and plug in our initial concentrations which we know what those values are because we solve for all of them. So for our initial concentration of conjugate base we set it is 0. 0. 8056 Mohler. Our concentration for hydro knee um is 0. eight six 3704 Moller. And our concentration For our product is zero 0863704. Okay and our change in concentration is we're going to subtract our concentration of hydro ni. Um So this is going to be because it is the smaller value. So we have minus 0. -0.0863704. And then on this side it's going to be plus because our reaction is moving in The forward direction. Okay? So plus zero 0.0 eight. Zero point excuse me. Zero point 0863704. Okay. And so after the reaction, what we're going to have And the Equilibrium Road is 0.0116864. We're gonna have zero here and then we're going to have 0.863704. So now that our ice table is complete, we can use the Henderson Hasselbach equation to software P. H. Because we're working with a weak acid and its conjugate base. The Henderson Hasselbach equation is what we use to software P. H. So ph the equation is P. H. Equals P. K. A. Plus the log of the conjugate base. The concentration of the conjugate base over the concentration of the weak acid. So let's go ahead and replace what are conjugate base and weak acid are. Okay so our conjugate base over our weak acid. Okay. And we know what the values are. So for P. K. A. We know what the P. K. Is. We know what the concentration is for the conjugate base and we know what the concentration is for the weak acids. So now we just need to plug these values in. So for P. H equal to P. K. A. We were told, is 7.21 And this is going to be plus the log of the conjugate base. And now we know what the value is because we use the ice table to find out our concentration at equilibrium. So for the conjugate base it is 0. Divided by. And our concentration for the weak acid is 0.0863704. Okay, so what we'll get is Ph. is equal to 6.34. And this is our final answer. That is it for this problem. I hope this was helpful.