In this question where we are told that codeine and then we’re given the formula for codeine which is a commonly prescribed pain killer is a weak base a saturated aqueous solution of codeine contains one gram of codeine in 120 mils of solution and it has a ph of 9.8 given this information we’re asked to calculate what is the kb of codeine so what i’m going to do is

I’m going to indicate that we have codeine this base using the symbol b and it’s a base so i need to set up the k b reaction to figure out the ph so there we are just like that and bases of course make all h minus so whenever we’re doing a i stable we’re going to be working in moles per liter so the first thing i’ve got to do is figure out how many moles per liter

Of the codeine that i’ve got so it doesn’t give me it doesn’t give molarity it gives me grams in 120 mils so we’re going to go ahead and calculate the moles of codeine and that’s always going to be equal to your grams over your molar mass and so i’m gonna have to go ahead and calculate the molar mass so i’ve got 18 times 12 point oh one for all of the carbons plus

21 times one point oh one for the hydrogen’s plus fourteen point oh one for the nitrogen and then 3 times 16 for the oxygens and that ends up being two hundred ninety nine point four okay so once i’ve done that map 1 divided by that for that ends up giving me 3 point 3 4 by 10 to the minus 3 moles okay but what we want to do is we need to divide that by the leaders

Of solution in order to get the molarity so i’ve got 120 mils or 120 by 10 to minus 3 liters and so my molarity if it’s going to be the answer that i got before divided by 0.1 2 and then that ends up being 0.02 7 8 molar so that becomes my initial concentration of my codeine 0.02 7 8 this is irrelevant and these are 0 and 0 before it becomes before it comes into

Equilibrium ok so now let’s go through and see if we can solve well given the ph it’s not very helpful because what we really need is the equilibrium rh – but if i do 14 minus the ph i can get the poh so 14 minus 9.8 is gonna give me 4.2 i for the ph and i know that the the oah – is really equal to 10 to the minus poh which would be 10 to the minus 4.2 i so when

I do that i get my equilibrium ohy concentration which was 6.3 1 by 10 to the minus 5 so at equilibrium the concentration of our h- is 6.3 1 by 10 to the minus 5 so what’s going to happen is this comes to equilibrium it always moves towards the zero so this will go down by x so giving me point zero two seven eight minus x this will go up by x this will go up by

X and so what we know is that x equals six point three one by 10 to the minus five so now i can go ahead and calculate my kb so my kb is druce my concentration of rh – because my x in this case multiplied by the concentration of my conjugate acid ph plus again which is just x multiplied by my equilibrium concentration of my base which was my point zero two seven

Eight minus x so let’s put all of that in these two guys are the same the hydroxide and the bh+ of the same so i’m just going to put that in is six point three one five by ten to the minus five squared divided by the concentration of my base point zero two seven eight – six point three one by ten to the minus five it’s a weak base so it should be a small kb let’s

See what we get and i ended up getting one point four four by ten to the minus seven kb’s don’t need any units and that’s definitely a small number

Transcribed from video

Kb of codeine By Lee West