- Acid–base titrations
- Worked example: Determining solute concentration by acid–base titration
- Titration of a strong acid with a strong base
- Titration of a strong acid with a strong base (continued)
- Titration of a weak acid with a strong base
- Titration of a weak acid with a strong base (continued)
- Titration of a weak base with a strong acid
- Titration of a weak base with a strong acid (continued)
- 2015 AP Chemistry free response 3b
- 2015 AP Chemistry free response 3c
- 2015 AP Chemistry free response 3d
- 2015 AP Chemistry free response 3e
- 2015 AP Chemistry free response 3f
- Titration curves and acid-base indicators
- Redox titrations
Titration of a strong acid with a strong base (continued)
Calculating the pH at and after the equivalence point for titration of strong acid, hydrochloric acid, with strong base, NaOH. Created by Jay.
Want to join the conversation?
- Is it right to use the Hasselbach equation pH= pKa + log([conj. base]/[acid]) ?(11 votes)
- Nope! Since we are mixing a strong base with a strong acid, at no point do we have a buffer (which would require a weak acid plus its conjugate base, or a weak base plus its conjugate acid). Since we don't have a buffer we can't use the Henderson-Hasselbach equation.(39 votes)
- What is the pH level of soil from the Sahara desert?(0 votes)
- Why does the pH increase so much from only ,0002L of the hydroxide ion solution?(4 votes)
- Remember that pH is a log scale so the change will "appear" to happen fast because each unit is changing by a factor of 10. In a titration what you are basically doing is trying to find the number of either H+ or OH- in the solution by adding a known concentration of the opposite so at pH 7 the number of OH- and H+ are equal, the solution is neutral. Once you hit that point a small change in the amount of OH- will rapidly increase the pH because the solution you're adding is very concentrated and there is no H+ left to react with the OH-'s you're adding.(9 votes)
- Why does Dave insert the negative sign on the top of the O instead of at the end of OH?(4 votes)
- It doesn't really matter where you place the negative charge on the OH species. OH is so strongly held together that the negative charge belongs to the OH species and not to any individual atom (the O will be more polar)(4 votes)
- I was wondering why we couldnt use MV=MV in this situation? For the past videos I have been able to use MV=MV to find the concentration and use the pH equation to find pH. With this problem, I set it up as (20)(.5)=(x)(40) and got x=.25M
The -log(.25) is not 7. Can you please tell me why MV=MV would not work to get the answer for this question?(3 votes)
- Several problems:
The MV=MV formula measures the original, not the final values. You are being asked the final amount.
You are mixing and matching units. You must use liters, not mL for pH computations.
Since you are mixing together the two substances, the final volume will be the sum of the two volumes.
The chemicals are reacting, so you no longer have what you started with. The H⁺ and OH⁻ ions are combining with each other, so that they no longer exist.
You are not taking into account the amount of H⁺ that naturally exists in water.(9 votes)
- What does the equivilance pt. mean?(2 votes)
- The equivalence point is where the moles of acid, and the moles of base in a titration are equal to each other in the solution.(4 votes)
- Is there any way to calculate mathemetically that the pH becomes 7 when 20 ml of NaOH neutralizes 20 ml of HCl?(2 votes)
- When you have added exactly the same number of moles of NaOH as there are moles of HCl, then they will have reacted together to give water. So you have no excess acid, nor any excess base. All you have is the hydronium and hydroxide ions resulting from the autoionisation of water. Furthermore, these two concentrations will be equal to each other.
At 25 C, Kw = 1.0 x 10^-14. Kw = [H3O+][OH-] = [H3O+]^2 because the two ion concentrations are equal. Taking the square root, [H3O+] = 1.0 x 10^-7. Take the negative log of this and you find that the pH is 7.0.(3 votes)
- How do we know that all the hydrogen ions and hydroxide ions will react to form water? Shouldn't some of them dissociate back into hydrogen ions and hydroxide?(2 votes)
- There is an equilibrium between the (OH- & H+) and H2O. Some will continuously dissociate while others will reform water.(3 votes)
- I got pH=11.395773947 for problem d) which is slightly off. During my calculation, I didn't rounded up randomly. What caused this? If I write a slightly wrong answer like this on an exam, I am going to get it wrong.. right? How do I avoid slightly wrong answers like this?(2 votes)
- Don't worry - your answer was correct (rounded off to 11.396).
Jay rounded off prematurely (you should only round-off at the end, not during intermediate steps). I assume he did this because keeping three significant figures throughout would make his calculations very messy and harder to follow ...(3 votes)
- Is the pH at the equivalence point going to be 7 for all strong acid and bases?(2 votes)
- You are correct, equivalence point represent the completion of the neutralization reaction. Since both acid and base are strong, the salt is neutral.(2 votes)
- We've been looking at the titration curve for the titration of a strong acid, HCl, with a strong base, NaOH. In the previous video, we've already found the pH at two points on our titration curve, so we found the pH before we'd added any of our base, we found the pH at this point, and we also found the pH after we added 10 mls of our base, we found the pH at this point. This is Part A of our question, this is Part B of our question. Now we're on Part C. What is the pH after the addition of 20 mls of a .500 molar solution of sodium hydroxide? So how many moles of hydroxide ions are we adding to our original acid solution? We can figure that out by this concentration rate here. So the concentration of sodium hydroxide is .500 molar. That's also the concentration of hydroxide ions in solution, since we're talking about a strong base, so the concentration of hydroxide ions in solution is equal to .500 molar. Remember: molarity is moles over liters. So this would be equal to moles over liters. How many liters do we have? We have 20 milliliters, so 20 milliliters would be equal to, move our decimal place, one, two, three; that's .02 liters. So to find how many moles, just multiply .5 by .02 and you will get .01 So we have .01 moles of hydroxide ions. So that's how many moles of hydroxide ions we're adding to our original acid solution. In the previous video, we already calculated the moles of hydronium ions in solution, and it was the same number as this: .0100 moles of H3O+ So you could watch the previous video, or you can just look at, it's the same numbers: 20 milliliters here, 20 milliliters here, .500 molar here, .500 molar here. So it's the same calculation to give us moles of H3O+, hydronium, or you could just consider that to be moles of H+. So we have an equal number of moles of base as we do of acid, and the base that we're adding is going to neutralize the acid that's present. So the H3O+, the hydronium that's present, is neutralized by the base that we add. And so H3O+ donates a proton to OH-, so we get H2O, and then if H3O+ donates a proton, we also get another molecule of H2O. So we get two molecules of H2O here, and we are starting with .01 moles of OH- So let's go ahead and write that in here. So we have: .01 moles of base, and that's the same number of moles of acid that we have. So .01 moles of acids. So, this time, we have enough base to completely neutralize our acids. So everything is reacting in a 1:1 ratio here. So all of our base is going to react, and it's going to completely neutralize our acids. So when that happens, the pH should be just the pH of water. The pH of our solution should be the pH of water, which we know is equal to seven, so 7.00 I could have written this another way. I could have written: HCl + NaOH, right? This would give us, this would give us H2O, 'cause H+ and OH- give us H2O, and then we would have NaCl left, right? We would have a solution of sodium chloride. So, an aqueous solution of sodium chloride. And, if our acid and our base completely neutralize each other, we're just left with an aqueous solution of sodium chloride, and so the pH is just the pH of water, because sodium and ions and chloride anions don't interact with water enough to change the pH. So the pH, after we've added 20 mls of our base, is equal to seven. So we can find that here on our titration curve. So 20 mls of base added, the pH should be seven, so we can find this point on our titration curve. This is the equivalence point. So let me go ahead and draw a line down here. So right here is our equivalence point. Let me write that. So, our equivalence point has been reached. We've added enough moles of base to completely neutralize the acid present, so we've reached the equivalence point. Finally, in Part D, they want us to find the pH after the addition of 20.2 mls of a .500 molar solution of sodium hydroxide. So just like before, we need to find the moles of hydroxide ions that we're adding. And so the concentration of hydroxide ions is, once again, .5, so .500 molar. Molarity is moles over liters, so we wanna find moles. How many liters are we adding? Well, 20.20 mls, is the same thing as .02020 liters. So, we just need to solve for moles; and you can probably do this in your head. I'm just going to use the calculator here to show you the answer. So: .5 x .02020, gives us .0101 So that's how many moles of hydroxide ions we have: .0101 moles of hydroxide ions. Okay, remember: the hydroxide ions reacted with the hydronium ions. We talked about the fact that H3O+ plus OH- gives us 2H2O. This time, we're starting with .0101 moles of hydroxide ions. So let's write: .0101 moles. And hydronium, we only started with .01 moles of hydronium, so, .01 moles of hydronium ions. And so, this time we have more base than we do acid, so all of the acid is going to be neutralized, right? So the acid is going to be completely neutralized. We're gonna be left with nothing. So all the acid is gone, and most of the base is going to react. So we're gonna get the same, we're gonna lose the same amount of base, so we're gonna lose .0100 moles of base, and so we're left with a very small amount of base. We're left with .0001 moles of base left over. So all of the acid has been completely neutralized and we have a small amount of base left over. Next, let's think about the total volume. We started with 20 mls of our acid solution, right? We started our titration with 20 mls, and at this point of the titration, we've added 20.2 mls more. So we've added 20.2 mls more. This gives us a total volume of 40.20 mls. So we have 40.20 mls here. And now we can calculate the concentration of hydroxide ions in solution. So what's the concentration of hydroxide ions? Concentration is moles over liters, and so we have .0001 moles of hydroxide, and the volume would be, this is in milliliters, so that's the same thing as .04020 liters. So we can go ahead and do that calculation. Take out the calculator here. So we have .0001 divided by .04020, and we get .002 So our concentration of hydroxide ions is equal to .002 molar. Our goal was to find the pH, but right now, we just have to find the pOH. So, the pOH is equal to the negative log of the concentration of hydroxide ions. So this is the negative log of .002 So: -log(.002) Let's see what that gives us on the calculator. - log(.002) gives us a pOH of 2.7 So the pOH is equal to 2.7 And finally, to find the pH, we need to know one more equation. The pH plus the pOH is equal to 14. So if we plug in 2.7 into here, the pH is equal to 14 - 2.7, which is, of course, 11.3 So we finally found the pH. So let's think about where this point is on our titration curve. We've added 20.20 mls of our base, let me go ahead and use blue for this so we can see it a little better, we've added 20.20 mls of our base, and the pH, we just found to be 11.3 So we're just barely past this 20 here. It's really hard to draw a straight line. I'm not doing a very good job. But our pH should be 11.3 so that allows us to find this point on our titration curve, so somewhere around there.