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Ch.16 - Aqueous Equilibria: Acids & Bases

Chapter 16, Problem 94

A vitamin C tablet containing 250 mg of ascorbic acid 1C6H8O6; Ka = 8.0 * 10-52 is dissolved in a 250 mL glass of water. What is the pH of the solution?

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Hi everyone for this problem. It reads in a 250 millimeter glass of water, an aspirin that contains 75 mg of a seagull silic acid with a K. Equal to 3.3 times 10 to the negative four is dissolved, calculate the ph of the solution. Okay, So because we're dealing with an acid here, we'll be able to calculate the ph of the solution by first setting up an ice table and finding out the hydro ni um ion concentration. Okay, so we want to find out the concentration of hydro knee um ions. And like I said, we're going to need to set up an ice table to do this. So let's go ahead and write a reaction for this. So the reaction that we're going to write a single silic acid is a weak acid and it's going to dissociate in water as follows. We're going to let a seat silic acid, B H A. Okay. And it's going to react with water and at equilibrium this is going to produce we're going to let the contract get base of acero silic acid be represented by a minus. And this is also going to produce hydro ni um ions. Okay, so with our ice table we need to set it up where we put ice on the side and this represents, I represents the initial concentration and we need to solve for that. Okay, so let's go ahead and calculate the initial concentration of aceto silic acid. The first thing that we're going to want to do is find out our moles of a seed of silic acid in the problem. What are we given And what do we know? We're starting with We know we're starting with 75 mg of acetyl silic acid. So let's start there. We have 75 mg. Okay? And we want to go from milligrams, two moles of acetyl silic acid. So the first thing we're going to want to do is go from milligrams to grams. Okay, And in one graham there is 1000 mg. Now our units for milligrams cancel. And we're left with units of moles. Now we want our grams. Excuse me. Now we want to go from grams to moles. Okay. And one mole of a seagull silic acid. The molar mass is 180.158 g. Will need to use our periodic table to find out the molar mass of aceto silic acid. The units of grams cancel. And we're left with moles. Okay, so now we can solve for moles of acetate silic acid. And when we do this, what we get is 4.1630 times 10 to the negative four moles. Okay, But we need concentration and our Iroh of the ice table. This represents concentration and remember concentration is moles over leader. So our concentration for cedo silic acid which were having represented by H A. Here is going to equal moles over leader. We just solve for moles? We said the moles is 4. times 10 to the negative four. Now the leaders or the volume we're told in the problem is 250 mL. We need this in leaders. Okay, so we're going to divide 250 by 1000. And this is going to give us 0.250 liters. So when we solve then for the concentration, what we're going to get is 1.6652 times 10 to the negative three molar. This is the concentration we're going to plug in to our ice table. So let's go ahead and do that now. So our initial concentration for aceto silic acid is one point 52 times 10 to the negative three. Okay. And we don't include the water. So we'll just go ahead and not put anything there and we have zero product. Okay, so for our ice table, I like to differentiate our products from our reactant and we don't include liquids or solids. Alright, so that's why we have the water crossed off. So for our change in concentration because we have no products, our reaction is moving in the four directions. So that means our concentration of reactant is going to decrease and our concentration of products is going to increase. So for a change we have a minus X. We only have one mole of that. So our coefficient is one and then we have products forming. So this becomes plus X. So when we combine both rows of our equilibrium table or both rows of our ice chart. So we get 1.6652 times 10 to the negative three minus x. And then we get X. And X. So we need to solve for X. Because this value of X is what's going to give us the the equilibrium concentrations and essentially we want to know what X. Is for the hydro knee um ion concentration because that is what we're going to use to figure out the ph Alright, so our goal now is to solve for X. And we're going to calculate the hydro ni um ion concentration using the equilibrium constant expression so that K. A. Okay, so we get K A. Is equal to the concentration of products. This expression is the concentration of products over the concentration of reactant. See that's the expression for equilibrium constant. And what we're going to do is we're going to plug in the equilibrium row of our ice table into this. So we get the concentration of products is going to equal X times X. And the concentration of reactant is 1.6652 Times 10 to the -3. Okay and now we're going to so when we simplify this it becomes K A. Is equal to X squared over 1.6652 times 10 to the negative three. And in the problem we're given the K. A value, we're told the K. A value is equal to 3.3 times to the negative four. Okay So then what that means is we can replace this K. A. value with its actual value which is 3. times 10. So then I get four. Alright so now we're going to determine if we can remove X. The way we're going to do this is we're going to divide the initial concentration by the K. A. Value. If this value is less than 500 we can't remove X. But if it's greater than 500 we can remove it. So the The initial concentration over H. A. is less than 500. So let's go ahead and write that. So X. is not negligible. X. Is not negligible because the the initial concentration over the K. A. Which our initial concentration is the 1.6652 times 10 to the negative three. This value is less than 500. Okay? So that means we need to leave X there and solve for it. All right, so let's go ahead and simplify this. Let's get rid of our fraction. And when we get rid of our fraction, what we're going to get is X squared is equal to 3.3 times 10 to the negative four times 1.6652 times 10 to the negative three. Okay so we get X squared is equal to we have to foil. Excuse me? I did not write that correctly. We should have a minus X here. Okay, so this all should say minus X. Because we're copying what's on the equilibrium row of the ice table. So we need to foil this out. Okay, when we're simplifying here, so what we're going to get then is 5.9516 times 10 to the negative seven minus three .3 times 10 to the -4 X. So we want to bring everything over to one side. Okay? So when we do that, what we're going in we're gonna set it equal to zero. So X squared plus 3.3 times 10 to the negative four X minus 5. 516 times 10 to the negative seven equals zero. So now what we're going to do is we're going to solve for X. Using the quadratic formula. Mhm. Okay. And the quadratic formula is X. Is equal to negative B plus and minus the square root of B squared minus four A. C. All over two. A. And when we look at our equation right here, what we can say is A is equal to one, B. is equal to 3.3 times 10 to the - and C is equal to 5. Negative excuse me equal to negative 5. 516 times 10 to the negative seven. So we have everything we need to plug into our quadratic formula. So let's go ahead and do that so we can solve for X. And we're going to get two values for X here because we have a plus or minus, we're going to get two values. So let's go ahead and plug in. So we get X. Is equal to negative 3.3 times 10 to the negative four plus and minus the square root of b squared just 3.3 times 10 to the negative four Squared -4 times a times c. Yeah. Okay so this is all over two a. So this is all over two times one. So when we solve this we're gonna get a positive and negative value for X. So for the positive value we're going to get 5. 35 times 10 to the -4. And we're going to get a negative value of negative 9. times 10 to the -4. Okay, so we need to figure out which value for X. We're going to use since we have to we will use the positive value for X because we can't have a negative concentration. Alright. And so based off of our ice table recall that X equals the concentration of hydro knee um ions. And the value that we have for X. Is 5.94435 times 10 to the negative four moller. So the last thing that we're going to do is calculate the P. H. And we're going to calculate the ph because we know that P. H. Is equal to the negative log of hydro knee um ion concentration and now we know what the hydro knee um ion concentration is because we saw for it it was the value for X. Based off of our ice table. Okay so we'll just go ahead and plug that value in. So we're gonna take the negative log of 5.94435 times 10 to the negative four. And what p what we're going to get for D. P. H. Is P. H. is equal to 3.23. And that is our final answer. That is it for this problem? I hope this was helpful.