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I've taken this problem from Chapter four of the chemistry and chemical reactivity book by Kotz treichel and Townsend and I've done it with their permission so let's do this this example a one point zero three four gram sample of impure oxalic acid is dissolved in water in an acid-base indicator with and an acid-base indicator added the sample requires thirty-four point four seven milliliters of zero point four eight five molar sodium hydroxide to reach the equivalence point what is the mass of oxalic acid and what is its mass percent in the sample so before we even break into the math of this let's just think about what's happening we have some oxalic acid which looks like this it's really kind of two carboxylic acid groups joined together if that means anything to you watch the organic chemistry playlist if you want to learn more about that so we have a double bond to one oxygen and then another bond to a hydroxide and we have that on the other carbon as well this is oxalic this right here is what oxalic acid is and it's an interesting acid because it can actually donate two protons this proton can be nabbed off and this proton can also be contributed and it's actually resonance stabilized if that doesn't mean anything to you don't worry you'll learn more about that in organic chemistry but the important thing to realize here is that there's two protons to nab off of it there are two protons to nab off of it now each molecule of sodium hydroxide or look we should when you when you put it in the water really just dissolves and you can really just think of it as hydroxide each molecule of hydroxide can nab one of the hydrogen protons so for every one molecule so for every one molecule of oxalic acid you're going to need let me just for every one molecule you're going to need two hydroxides one to nab this hydrogen proton and then another one to nab that hydrogen proton so let's just draw the balanced let's write down the balanced equation that we're dealing with here so we're going to start off with some oxalic acid so that has two hydrogen's so it's h2 two carbons and then four oxygens oh four it's dissolved in water so it's an aqueous solution and to that we're going to add sodium hydroxide sodium sodium hydroxide now I just told you that you're going to need two of the hydroxides to fully neutralize the oxalic acid so you're going to need two of them and this is also in our aqueous solution and once the reaction happens this guy will have lost both of the hydrogen protons so let me draw that so it will look like this no more hydrogen so it's C 2 O 4 it'll have a negative 2 charge and actually it will you could imagine that it might be attracted to this positively charged sodium and two sodium's in particular so this has a negative 2 charge we could even write it there if you want to - and then you could have the sodium's over here you have these two sodium's that have a 2 plus this entire molecule becomes neutral they are attracted to each other they are still they are still in an aqueous solution and then the hydroxide NAB's the protons and then you are left with just water so plus 2 moles plus or 2 molecules depending on how we're viewing this plus plus 2 waters we could write it I'll just use that same orange color plus 2 h2o s one of these hydrogen's are coming or one of the one of the hydrogen's in each of the water molecules are coming from the oxalic acid and so two of these hydrogen's in these two moles of the water are coming from one entire molecule of oxalic acid now let's actually do the math we have 34 point 4 7 milliliters of the solution that has the sodium hydroxide and I'm just going to convert that to liters just so it's easier to deal with the molarity right over there so we have 34 0.47 milliliters milliliters we could write it of this solute we understand that that's the case so let's just multiply or actually divide so this is x this is x we have 1 liter 1 liter for every thousand for every thousand milliliters and then this will give us the milliliters cancel out thirty-four point four seven divided by a thousand is zero point zero three four for seven liters of this 0.485 molar sodium hydroxide solution so let's figure out how many actual molecules of sodium hydroxide we have this is the solution and we know it's concentration it's zero point four eight five molar so let me do that in a different color zero point four eight five molar this information allows us to figure out the actual molecules of sodium hydroxide so we want to multiply this by we have we have point we have zero point four eight five moles moles of sodium hydroxide for every for every one liter of this solution that's what that that's what the molarity tells us we have point four eight five moles per liter so the liters cancel out and then now we're going to actually have to get a calculator out and this will tell us how many moles of sodium hydroxide we have in this solution so let me get my my calculator there we go all right let me just multiply these two numbers so we have point zero three four four seven times times point four eight five is equal to let me put this down here zero point one six seven and we only have actually with the we only have three significant digits here so we're going around to three significant digits so we'll just go with is zero point zero one six seven so let me let me move that over off the screen so this is going to be equal to let me this is going to be equal to zero point zero one six seven and all we have left here are moles of sodium hydroxide moles of sodium hydroxide now what we want to do is we know that this many moles of sodium hydroxide are going to completely react with however many moles of oxalic acid we have now we know that we need two moles of this for every mole of oxalic acid or for every mole of oxalic acid that that completely reacts we need two moles of this so let's write that down let's write this down in a new color so times we need two moles to moles of sodium hydroxide we got that from our balanced equation right there and you can kind of obvious it needs this it needs one mold or one molecule will take this proton and then you need another molecule to take that proton so we need two moles of sodium hydroxide for every one mole for every one mole of oxalic acid for every one mole of h2 C 204 so we essentially are just going to divide this number by two let me get the calculator back so we're just going to divide point zero one six seven divided by two divided by two once again three significant digits point zero zero eight three five so this is going to be zero point zero zero eight three five moles of oxalic acid h2co3 to figure out the mass and we know the molar mass of oxalic acid we know that carbon let me write these down we know that hydrogen has a molar mass let me let me write it this way molar mass molar mass if you have a mole of hydrogen it has a molar mass of one gram if you have and this just comes from its atomic weight if you have carbon its molar mass if you have a mole of carbon its molar mass is twelve grams and if you have oxygen its molar mass molar mass is 16 grams so what's the molar max of oxalic acid well we have two hydrogen's so that's going to be 2 grams right 2 times 1 grams that's the hydrogen's there we have 2 carbons so it's going to be plus 24 grams 12 grams for each of these carbons and then we're going to have 4 oxygens that weigh in if we have a mole of them at 16 grams so that's going to be plus 64 so what does this come out to 24 plus 64 24 plus 64 is 88 is 88 right 2 plus 6 is 8 right it's 88 plus 2 more is 90 grams is 90 grams so it has a Sox aaalac acid if you had a mole of oxalic acid it would be 90 grams so we could say 90 grams per mole of C 2 C 2 I should write the H first of H 2 C 2 O 4 so let's go go back to the math here so we had I'll rewrite it over here we had we know we're dealing with 0.0083 5 moles moles of oxalic acid h2co3 MS let me do this in a different color this color is getting monotonous we know that there are 90 grams of h2 C 2 o 4 for every mole for every 1 mole of h2 C 2 0 for this is its molar mass so now we just multiply this number and we'll figure out the grams of oxalic acid that and that cancels out and then we just take the number that we had and multiply it by 90 so times 90 this just says the previous answer which is the number of moles of oxalic acid times its molar mass will tell us the grams of oxalic acid so we get point seven five I'll just round it to two since we only have three significant digits and actually well this even the 90 isn't exact so it's a little bit it's a little bit but I'll just round to three significant digits so seven point seven five two so this is equal to this is equal to zero point seven five two grams of oxalic acid h2co3 evacs aaalac acid we've just answered it that answer right there is zero point seven five two grams now the next part is is what is its mass percent in the sample well the sample of impure oxalic acid right over here was one point zero three four grams so we just have to say what percentage is 0.75 to of 1.0 three four so let's get the calculator back so we have the point seven five two divided by one point zero three four and we get seventy two point seven percent seventy two point seven percent so the second answer the answer to the second part right over there is seventy two point seven percent we were able to figure out that this impure oxalic acid sample is seventy two point seven percent actual oxalic acid

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