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Current time:0:00Total duration:10:12

Video transcript

let's say we're doing a titration and we start with 20 MLS of 0.5 0 0 molar HCl so we're starting with a strong acid and to the strong acid we're going to add a solution of a strong base we're going to add a point 5 0 0 molar solution of NaOH and as we add the base the pH is going to increase and we can show this on our titration curve so we put the pH on the y-axis and on the x-axis we put the volume of base that we are adding so in Part A our goal is to find the pH before we've added any of our sodium hydroxide so the addition of zero point zero MLS of our base so if we haven't added any base the only thing we have present is acid right we know that HCl is a strong acid so it is going to on eyes 100% so if HCl donates a proton to h2o then we get h3o plus and we would get the conjugate base to HCl which is CL minus if we're starting with a concentration of 0.5 0 0 molar HCl let's go ahead and write that point five zero zero molar HCL and since we know this is 100 percent ionization because we have a strong acid that's the same concentration of hydronium ions that we'll have in solution so we have 0.5 0 0 molar for the concentration of hydronium ions and now it's easy to find the pH because we know that the pH is equal to the negative log of the concentration of hydronium ions so let's go ahead and do the calculation all right we take the negative log of the concentration of hydronium which is 0.5 0 0 so that gives us point 3 0 1 so we plug this into here and we get a pH equal to zero point three zero one so we can find that point on our titration curve so it's right here at the beginning right we've added zero point zero MLS of base and our pH looks like it's just above zero here on our titration curve and we calculated it two be point three zero one all right let's find another point on our titration curve so this time after we add 10 ml of our base so we find 10 MLS of our base right here and we're trying to find this point right so what is the what is the pH after we add 10 MLS right well it looks like it's pretty close to 1 and let's see if we can calculate what the pH is all right so if we're adding base we know that the base that we're adding right the hydroxide ions that we're adding are going to neutralize the hydronium ions that are already present so first let's calculate how many moles of hydronium ions that we had present here so the concentration of hydronium ions is 0.5 0 0 so let's get some more room down here we know the concentration of hydronium ions is equal to point five zero zero molar and we know that concentration right or molarity is equal to moles over liters so this is the molarity that's equal to the moles of h3o plus over the liters so what was the original volume of acid that we started with let's go back up here to our problem we started with 20 milliliters all right so let's write that right here 20.00 MLS if I want to convert that into liters I move my decimal place 3 to the left one two three so that's point zero two liters so let me go ahead and write that in here so we have we have point zero two liters and they're going a lot of zeros in this video and it's going to be it would be annoying if I said them all so I'm just going to say point zero two and not not list all of those zeros that we have all right so to solve for moles we just multiply 0.5 five point zero two and so we would come out with the moles of h3o plus and this is equal to point zero one moles of h3o plus so that's the moles of acid that we're starting with and we're going to add some bass right we're adding some sodium hydroxide and we know sodium hydroxide is a strong base so the concentration of sodium hydroxide is the same as the concentration of hydroxide ions right na plus and Oh H - so 0.5 0 0 molar solution of sodium hydroxide is 0.5 0 0 molar for hydroxide ions so we write down here the concentration of hydroxide ions is equal to 0.5 0 0 molar and once again we need to figure out how many moles of hydroxide ions that we have and what's the volume right we have 10 milliliters so in liters that would be move our decimal place 3 all right so that's point zero one so this is equal to point zero one for our liters solve for moles all right so we would just multiply 0.5 by point zero one and we would get our moles equal to point zero zero five all right so that's how many moles of hydroxide ions we have alright now that we've found moles of both our acid and our base we can think about the neutralization reaction that occurs right the base that we add is going to neutralize the acid that we had present so we had hydronium ions present and we added some base so the acid donates a proton to the base if OS minus picks up an H+ then we get h2o if h3o plus donates a proton we get another molecule of h2o so we end up with 2 h2o over here or if you prefer instead of writing h3o plus you could have just written h+ alright as your proton and your proton reacts with hydroxide to give you water so either way of representing the neutralization reaction is fine so let's plug in let's plug in our moles here so we know that we started with point zero one moles of h3o plus so let me write that down here point zero one moles of h3o plus and we started and we added I should say point zero zero five moles of hydroxide all right so we're we added point zero zero five moles of hydroxide ions all of the hydroxide is going to react right it's going to neutralize the same amount of hydronium ions so we're going to lose all of our hydroxide ions because the base is going to completely react with our acid so we're left with zero moles of our base alright so for losing that much hydroxides right we're also going to lose that much hydronium so that much is reacting with the hydroxide ion so we're losing the same amount of hydronium ions all right so Oh point zero one minus point zero zero five is of course equal to point zero zero five so that's how many moles of hydronium are left over right so we've neutralized half of the hydronium ions present and so we have another half left over all right so one half of the acid has been neutralized so one half of the acid is left let's think about the new concentration of hydronium ions in solution so the concentration of hydronium is equal to moles over liters alright so that would be point zero zero five moles of h3o plus well what is the new volume we started with 20 MLS of our acid and two that we added 10 ml of our base right so right up here we added 10 MLS of our base so the new volume is 30 MLS right we added 10 MLS and so our new volume is 30 MLS so to find the new concentration of hydronium ions we need to use 30 ml 0.03 liters right one two three so this would be point zero three liters and we can get our concentrations so let's get out the calculator here and let's take point zero zero five I'm going to divide that by point zero three and we get our concentration of hydronium ions to be 10.17 right so our concentration is equal to point one seven molar now that we know our concentration of hydronium ions right so our concentration is 0.17 we can find the pH because of course the pH is equal to the negative log of the concentration of hydronium ions so the negative log of 0.1 7 so we get on our calculator one more time and we find the negative log of 0.1 7 and that gives us a pH of 0.77 so our pH is equal to our pH is equal to 0.77 so now now we know we add 10 ml of base and the pH is 0.7 7 so let's look at go back to our titration curve up here alright and find that point we talked about earlier so we found this point right here after 10 MLS of base this pH right here is point 7 7 all right so we just did the calculation to show it on the next video we'll analyze this titration curve we'll look at some more points on it