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Current time:0:00Total duration:7:50

AP.Chem:

SPQ‑4 (EU)

, SPQ‑4.B (LO)

, SPQ‑4.B.1 (EK)

let's do another titration problem and once again our goal is to find the concentration of an acidic solution so we have twenty point zero milliliters of HCl and this time instead of using using sodium hydroxide we're going to use barium hydroxide and it takes twenty seven point four milliliters of a point zero one five four molar solution of barium hydroxide to completely neutralize the acid that's present all right so let's start with what we know we know the concentration of barium hydroxide it's point zero one five four molar and we also know that molarity is equal to moles over liters all right so we have point zero one five four is equal to let's make moles x over liters twenty seven point four milliliters is point zero two seven four liters all right so that's point zero two seven four liters we solve for X and X of course represents the moles of barium hydroxide so let's get out the calculator here and let's do that so let's get some room over here so we take point zero one five four and we multiply that by point zero two seven four and that gives us this would be four point two two times ten to the negative fourth right so that's equal to zero point zero zero zero four to two moles of barium hydroxide alright next let's write the neutralization reaction so we have barium hydroxide alright reacts with HCL so barium hydroxide plus HCL gives us for our products we have h plus and O H - so that's h2o and then our other product this is barium two-plus right this is BA 2 + and over here we have CL minus 1 so we have BA 2 plus and Cl minus one so you could cross those over so BA CL 2 right so bacl to barium chloride as our other product here all right next we need to balance our equation right we need to balance the neutralization reaction here so let's start by looking at let's start by looking at the chlorines so over here on the left all right we have one chlorine on the right we have two so we need to put a two right here and now we have two chlorines on both sides next let's look at hydrogen's so on the left side we have two hydrogen's here and then we have two over here so we have four hydrogen's on the left on the right we have only two hydrogen's so we need to put a two here for this coefficient to give us four hydrogen's on the right so now we have four and we should be balanced right everything else should be balanced let's look at the mole ratio for barium hydroxide to HCl for every right is a one here so for every one mole of barium hydroxide we have two moles of HCl so we already calculated how many moles of barium hydroxide that we use in our titration right that's point zero zero zero four two two so therefore we had twice as many of HCL so we can multiply this number by two and we'd figure out how many moles of HCl we have alright or you could set up a proportion right so for talking about if we're talking about a ratio of barium hydroxide to HCl our mole ratio is one to two alright and our moles of barium hydroxide let me go ahead and use a different color here that's up that's up here that's zero point zero zero zero four to two moles of barium hydroxide our goal is to find how many moles of HCl all right we're present and so obviously you just need to multiply point zero zero zero four to two by two and so we get X is equal to zero point zero zero zero eight four four right that's how many moles of HCl we have at our equivalence point all right so finally we just have to calculate the concentration of our acid solution right let's go back up here so we can see we started with right so we started with 20 milliliters of HCl all right and 20 milliliters would be move our decimal place Oh point zero two zero zero liters so now we have moles all right we have moles and we have liters so we can calculate the concentration all right so the concentration of HCL in our original solution would be we had point zero zero zero eight for four moles all right divide that by liters that was point zero two zero zero liters right 20 milliliters is equal to point zero two zero zero liters and so we can do our calculation here so we can now we can take point zero zero zero eight four four and we can divide that by point zero two zero zero and we get for our answer here point zero four to two molar all right so the concentration of HCl is equal to point zero four to two molar and we are finally done right that's our concentration of our acid solution let's see what happens if you try to use MV is equal to mV that shortcut that we learned about in the last video so this would be MV is equal to M V and let's do the molarity of the base times the volume of the base is equal to the molarity the acid times the volume of the acid so for our base the concentration was point zero one five four molar and the volume of base that we used was twenty seven point four milliliters in our titration for the acid right we don't know what the molarity is that's where we're trying to find in the problem and the volume was 20.0 milliliters alright so let's do that calculation so trying to use the shortcut way point zero one five four times twenty seven point four gives us gives us that number divided by twenty all right so we at we get point zero two one one alright so for our answer for X we get point or zero point zero two one one molar and so you can see that's not the correct answer right here we have here we got a concentration that's half of the concentration that we got when we did it the longer way and so if you want to use if you want to use this shortcut way for this problem you would have to multiply by two alright so if you multiply your answer by two then you get the correct answer zero point zero four to two molar and a lot of times students have a hard time figuring out what you do right so where do you where do you multiply by two how do you do that of course you can figure it out by looking at your vast equation up here all right but it's it's tricky for a lot of students and so that's why the shortcut way isn't always isn't always the best way you can still use it if you understand how to use it right but it's a little bit better for these problems when your mole ratio is not one-to-one to go through the longer way and do it if you want to be absolutely sure that you're getting it correct