neutralization reactions today we're going to learn how to do calculation with neutralization reactions also we're going to learn what actually is a neutralization reaction you see here I have a a lmon a lmon is a type of an acid a neutralization reaction is a reaction between an acid and a base so three things about neutralization reactions one of them I just said it's a reaction between an acid and base on the first slot I had an acid I showed you a lumon on this slot I had a base soap is also a base the products of the reaction are water and a salt we'll see that in just a second and the last thing at the end of a neutralization reaction there's no acid or base left so let's do some calculations the first one hydrochloric acid and sodium hydroxide react so the complete molecular would be HCL plus NaOH gives you NAC so NAC is the salt so that's our sa l T that's our salt and then we have water so the products of an acid base reaction are water and a salt or if you want to write the net net ionic reaction cross out the spectat ions which would be chloride because it's aquous on both sides and sodium because it's aquous on both sides and that would leave you with this reaction a proton plus hydroxide gives you water so let's do a problem with this we got three problems today what is a volume of one molar hydrochloric acid that's needed to neutralize exact ly 25 M of3 5 m sodium hydroxide so for this what you're going to do is you're going to use information about sodium hydroxide to find something out about hydrochloric acid so the balanced reaction is the one we just saw so what you're going to do is you're going to use multiply the volume and the marity to first get moles so what when you do that you see liters will cancel out so we have volume times marity and notice instead of writing Big M I wrote moles over liters and that enables me to cancel lers of sodium hydroxide now I want to go from moles of sodium hydroxide to moles of hydrochloric acid we see in the balanced equation there's a one to1 ratio so the third step is going to be the mole ratio step so that looks like this I'm going to say one mole of sodium hydroxide over one Mo of hydrochloric acid now the last step asks for volume now to do this what we're going to do is take the marity and actually since we're we know marity is equal to moles that's m o over liters we know liters is equal to moles divided by marity so what I'm going to do is put the 0.1 on the bottom and liters on the top so the last thing we're going to do for this one is I've you this is the same problem but what I've done is I've put in I've divided here by the marity and see what happens the liters of sodium hydroxide cancels when I multiply by marity and then I use a mole ratio and it cancels out the moles of sodium hydroxide and then the last step is I divide by marity which means multiplying by the reciprocal and so moles of HCL cancel and then when I do the math for this I end up with 0875 L of 1 mol HCL or if you want to express it in milliliters it's 87.5 milliliters of hydrochloric acid so another problem that's fun next problem now this one's not a 1:1 ratio it's a 1:3 ratio what volume of 0.5 Mol sodium hydroxide is required to neutralize exactly 50.0 milliliters of12 molar phosphoric acid H3 P4 so here we have the balanced equation the important thing to see in this balanced equation is we have a 1:3 ratio so the first step is you want to multiply the volume times the marity so we see that here and the next step is you want to use the mole ratio for the balanced equation so the mole ratio in this balanced equation is 1: 3 and so for this step I'm going to say say 3 moles of sodium hydroxide over 1 mole of phosphoric acid so the phosphoric acid cancels and then in the last step it's going to look like this you say you you're multiplying by the reciprocal again because you're dividing by mity to get volume so your lers of phosphoric acid cancel uh and the next moles of phosphoric ACI as they cancel then last moles of sodium hydroxide cancel and when the moles of sodium hydroxide cancel you're left with your units which are liters of sodium hydroxide so 005 time12 * 3 / .5 gives us this 0306 l of5 m sodium hydroxide or just 30.6 milliliters of5 M sodium hydroxide one more problem now this last problem we're going to this last problem we're going to determine the concentration so it's going to be a little bit bit different what concentration of sulfuric acid neutralizes 47.5 M of .1 molar sodium hydroxide if 25 Millers of sulfuric acid is used so we're trying to find the concentration of sulfuric acid we have its volume but we need the number of moles so we're going to use the sodium hydroxide and its volume of concentration to find the moles of sric acid so here's our balanced equation notice the differ thing about this balanced equation is that we have notice the thing about this balanced equation is that we have a one to two ratio unlike the ratio we saw before so what you do is you take the volume and the concentration of sodium hydroxide and those will cancel out and now you have moles of sodium hydroxide now the difference here is we have a 1 to2 ratio so we say 2 moles of sodium hydroxide over one mole of sulfuric acid now the last thing is we just want to Simply so the moles of sodium hydroxide have canceled so the only thing we're left with at this point is moles of sulfuric acid to find marity we just need to put moles over liters so all we need to do is divide by this volume right here of air sulfuric acid so the way this last step should look is you just put one over the volume of the sulfuric acid so we've got 25 M of sulfuric acid we just change that to liters and put that right there so it's 0475 time .1 / 2 times or divided 025 and that would give us 0950 molar or you could also write as 0950 moles per liter sodium hydroxide