Oxycodone 1C18H21NO42, a narcotic analgesic, is a weak base
with pKb = 5.47. Calculate the pH and the concentrations of
all species present (C18H21NO4, HC18H21NO4
+ , H3O+ , and
OH-) in a 0.002 50 M oxycodone solution.

Question

Oxycodone 1C18H21NO42, a narcotic analgesic, is a weak base 
with pKb = 5.47. Calculate the pH and the concentrations of 
all species present (C18H21NO4, HC18H21NO4 
+ , H3O+ , and 
OH-) in a 0.002 50 M oxycodone solution.

Accepted Answer

Hi everyone for this problem it reads Pyro Ledeen. A common building block used in organic synthesis has a Pkb of 2.69 for a 0.41 molar solution of Pyro ladin determine the ph and the concentration of all species present. So the question that we want to answer here is the first is the ph and then the concentration of all species present. Okay, so in the problem we're given the PKB. Okay, so this is going to be an important piece of information and we can find out the KB from the PKB Given. So let's go ahead and do that first. So to find the KB, what we're going to do is it's going to equal 10 minus P K B. So that then is going to be 10 minus 2.69. So once we do this this gives us a value of 2.04 1, 74 times 10 to the negative three. So that is the value that we need for our KB, which is our based association constant. Okay. And now what we can do is we can create an ice table by writing out the reaction between Pyro Ledeen and water. So let's go ahead and do that here. So we're going to create an ice table. So we'll write the reaction. So we have Pyro ladin is going to react with water to produce the following ion plus hydroxide. So remember. So for our ice table we have I C. E. And for our ice table we don't include solids or liquids. So we're going to ignore this water here. Alright, so the iro of our ice table represents initial concentration. And so our initial concentration for pyro ladin was given in the problem. We're told it's 0.41 molar. So that is going to go here and we have no product. So these both are zero For the initial concentration this means our reaction is going to shift in the ford direction, which means our concentration of reactant is going to decrease and our concentration of products is going to increase. So we have minus X. For the change for our react dense and plus X for the change of our products. And we only have one mole Of everything. Okay, so now let's combine both. Rose 0.41 -1. And then we're going to bring these two xs down. So now what we're going to do is we're going to write R K. B expression and KB is going to equal the concentration of products over the concentration of reactant. This is our expression And we're we were given the value of K. B. This was given and we saw for this excuse me. And when we saw for this this value was 2. times 10 to the negative third. And this is going to equal our concentration of products over our concentration of reactant. So we're going to do is look at our ice table and plug in the equilibrium rose for our products and our constant our our reactant. So let's just define that here. So that means this is going to equal our concentration of product. So we have two products over our concentration of reactant. So we just have our Pyro Ledeen because remember we don't include solids or liquids in our ice table or in our expression. Okay, so this is going to be that expression. So what we're going to do here is we're going to replace these concentrations with what's in our ice table for those. So that becomes the concentration of X times X over The concentration of 0. -1. So, this part right here is what we're going to plug in. So that means that we have we're gonna simplify this two X squared over 0.41 minus X. Okay, so now what we need to do is solve for X. Because if we solve for X, let's look at our rice table, this X is going to give us our concentration of hydroxide and from our concentration of hydroxide, we're going to be able to solve for the ph Okay, and by also knowing X, we're going to be able to solve for the concentrations of all the species present. Okay, so let's go ahead and solve for X. One thing we can do is check to see if X is negligible. This X and the denominator right here. We need to see if it's negligible if it's negligible. We can ignore it. All right. And the way we check to see if X is negligible is we take the initial concentration which is 0.41 and we divide it by the KB which is 2.74174 times to the negative three. And if this value is greater than 500 we can omit that X. But if it's less than 500 we cannot omit the X. And in this case it's less than 500. So we have to keep this X here. Alright, so what we're going to do is we're going to keep the X. And we're going to filter it out so that we can set everything equal to zero. So when we solve this we're going to cross multiply. What we're going to get then is X squared plus 2. Times 10 to the -3. X -8.317. 8.37 113 times 10 to the negative four is equal to zero. With this. We're going to use the quadratic formula to solve for X. Okay, and the quadratic formula is X is equal to negative B plus or minus the square root of B squared minus four A. C. All over two. A. Okay, so looking at are our equation that we just set equal to zero. This is going to be A. This is going to be B. And this is going to be C. So let's go ahead and plug those values into our quadratic formula. So X. Is equal to negative B. Where b is 2.74174 times 10 to the negative three plus or minus the square root of B squared. So this is 2. times 10 to the negative three squared minus four times A. Where A is one times C. So this is going to be times negative 8.3711, 3 Times 10 to the -4. And this is all over two times 1. Okay, so when we saw for X we're going to get two values. So the first value we're going to get is X. Is equal to negative 0. and 0. three. So X cannot be a value a negative value because we can't have a negative concentration X should be a positive value. So we can eliminate this. So that means then we'll get an answer of X. Is equal to 0.2793. Yes moller. And this is the same thing as the concentration of hydroxide ions. And the concentration of our other product because based off of our ice table X equals both of these. Okay, so we just solved for the concentration of both hydroxide and So we'll write those concentrations are concentration of hydroxide is 0.02793 molar. And our concentration for the following is the same. Okay. And then we can now also solve for our concentration of Pyro Ledeen because based off of the ice table, the concentration of Pyro Ledeen is equal to 0. X. So that is 0.41 minus 0.2793. And that is gives us a answer of 0.3821 molar. So when you write this out properly we get the concentration of Pyro Ledeen is equal to zero 3821 Moeller. Okay so we answered the question of the concentration of all of the species. The last thing we need to answer is the ph okay so the way that we're going to solve for the P. H. Is we know that the concentration of hydro ni um is equal to one times 10 to the negative 14 divided by the concentration of hydroxide ions. So we know what the concentration of hydroxide ions is. So let's go ahead and plug that in. So our concentration of hydroxide is 0.02793. So that gives us a answer for the concentration of hydro knee um equal to 3.5803 times 10 to the negative 13 moller. And let's recall that P H. Is equal to negative log of hydro ni um concentration. Okay so the ph is going to equal negative log of 3. Times 10 to the -13 Moller. Okay, so our final answer then is P. H. Is equal to 12.45. And this is our final answer. So we answered the question of concentration of all the species and we answered the question of Ph that is it for this problem? I hope this was helpful.

Oxycodone 1C18H21NO42, a narcotic analgesic, is a weak base with pKb = 5.47. Calculate the pH and the concentrations of all species present (C18H21NO4, HC18H21NO4 + , H3O+ , and OH-) in a 0.002 50 M oxycodone solution.

Verified Solution

Key Concepts

Weak Bases and pKb

Henderson-Hasselbalch Equation

Equilibrium Concentrations