 So let's try this as our last problem of the day so we need to make a carbonate buffer of pH 10. How many grams of sodium carbonate must we add to 1.5 liters of freshly prepared 0.20 molar sodium bicarb to make the buffer? The Ka of the bicarbonate ion is 4.7 times 10 to the negative 11. So how do we do these? Well first things first we know remember whenever you have sodium ion that just is a spectator so we can kind of throw that out okay so first thing we have to do is write our buffer equation okay so let's write that so hopefully you guys can write you would have noticed no to write this I guess would you guys have known right now plus H2O and then what is that going to do so that's the acid so we're going to have CO3 2 minus there plus H3O plus A3 okay so we also know that well the Ka equation we can write that down Ka equals CO3 2 minus concentration H2O plus concentration divided by the HCO3 concentration so the pH here right would give us the H3O plus concentration we can get the H3O plus concentration from the pH is okay so let's do that so the H3O plus concentration is going to equal 10 to the negative pH like that so 10 to the negative 10.00 so 1.0 times 10 to the negative 10 is going to be the hydronium ion concentration so we need to figure out what is the constant because we're looking for the mass of sodium carbonate you guys know what sodium carbonate is so I know it's not written here but it's this thing with two sodium ions and A2CO3 that's what we're eventually looking for the mass of that so in order to eventually find the mass we first need to find the concentration of it so are you okay with thinking like that so we're going to rearrange this equation so we've got Ka right we've got sodium bicarb concentration we've got H3O plus concentration we're going to rearrange this equation to get the bicarb concentration so if you notice I did a nice table here in my head okay so bicarb 0.20 right concentration didn't change why do you get the Ka is so small down here H3O plus 4.7 4 molar or CO3 2 minus okay so that's molar okay not moles not grams so I'm going to change this units to moles of CO3 2 minus or 1 liter of solution right that's what molarity is you guys remember molarity say if you were going to multiply that well we've got a volume up here right so that's how many liters of solution we've got 1.5 liters like that everybody okay with me doing that going to cancel with that but that we're going to erase this top stuff here that would be the number of moles of CO3 2 minus would be 0.141 we'll keep that digit right now moles of CO3 2 minus but that's moles of CO3 2 minus right we want grams of sodium bicarb okay so for this right you remember this reaction this helps for every one mole of carbonate ions right we have one mole of sodium bicarb so that gives us moles of sodium bicarb that's not what we want right we wanted grams of sodium bicarb so on the bottom down here one mole sodium bicarb up here grams of sodium bicarb that's going to come from of course the periodic table and I did that earlier 10599 so that gives us when we cancel grams of sodium bicarb okay so when we multiply through 14.9 but we're only going to two stick things so one five grand any questions on this one can I shut it off okay