Like what you saw?
Create FREE Account and:
- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics
Tips for Making a Buffer via Neutralization - Concept
M.Ed.,San Francisco State Univ.
Jonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.
Here are some tips, and tricks for making a buffer solution. Remember a buffer has other two options. Number 1; it has to have a weak acid, and its conjugate base. Or you might see a conjugate salt. Or it has to have a weak base, and its conjugate acid or salt.
So here I'm going to show you two instances where you can make a weak acid, and its conjugate base. Or weak base, and its conjugate acid. So here to make a buffer, the first example that we have is you need excess weak acid that's neutralized by a strong base. The strong base needs to be the . So that's why you have excess weak acid.
So let's take a look here. We have acidic acid, CH3COH, and we add it to NaOH. Then so our products, we ended up with CH3COONa plus water. So we have a salt. The important part about this salt is the acetate, because that CH3COO-, when you put it in water, it dissociates. So this is actually the conjugate base of Acetic acid. So that's what you have here.
So we have molarity in volume. So since we're neutralizing, remember we need to take care of moles. So you multiply then to get moles. So we end up with 0.500mol of Acetic acid, and 0.250mol of NaOH. So this is what we have at the start. We have obviously no products so far. The water, we don't care about.
Then, since the ratio is 1 to 1 for mole ratios, we know that the NaOH is limiting. So 0.250mol gets used up of each of them, because that's a 1 to 1 ratio. So in the process, +0.25 moles of the conjugate base, or conjugate salt gets made. So that means that after we're done we have 0.250mol of Acetic acid. We have no Sodium Hydroxide left over.
In our particular solution, we have 0.250mol of CH3COONa which means what we're left with is, we're left with Acetic acid. We have conjugate base. So what we've done is we've made a buffer. Now keep in mind this was excess, this was limiting. The NaOH.
So the last thing you want to do is get the Molarity. So we take the 0.250mol and divide it by now a 1/4 volume. So we take the initial volumes of Acetic acid, and Sodium Hydroxide, and add them together. So 0.500 liters plus 0.250 liters gives us 0.750 liters.
So to get the molarity of the weak acid, and its conjugate base, you would just divide both of those. So we end up with 0.333 molar Acetic acid, CH3COOH. We end up with 0.333M CH3COO- which is the conjugate base.
So these two things when you put them together, they make a buffer solution. That's how you make a buffer with excess weak acid being neutralized by a strong base.
So now, instance number two would be on the flip side. We would have excess weak base that's neutralized by a strong acid. So basically it's flip flopped. So strong acid would be limiting. So we have a equation here HCl plus NH3. So we make NH4Cl, that's the product. Then this is all on water. So the water is the solution. So we make a salt here.
Now the important part, the NH4+ of that salt, because that's the conjugate acid of the weak base, NH3, Ammonia. So what we do is we do the same thing like before. So we multiply to get the number of moles. So we end up with 0.200mol here of HCl, and 0.375mol of NH3. We have none of the NH4Cl to start with.
Then we do same thing. So what's limiting? Well, the HCl is, because it's 1 to 1 ratio, and we have less of it. So 0.200mol gets used up. Then we make 0.200mol of NH4Cl. So we end up with 0.200mol of this. We end up with 0.175mol of Ammonia, and none other product.
Then after that, then all we have to do is calculate the Molarity. So we would divide by 0.4 plus 0.5 that's 0.9. So you do the Math. After you do the Math, you end up with 0.194M NH3. You do the Math and you end up with 0.222M of NH4+, the conjugate acid.
Once again, you have a weak base, and you have its conjugate acid. So together they make a buffer. So if you notice the pattern, the pattern is you have to have excess of your weak acid, or weak base. When it's neutralized by the strong opposite thing whether it's a base or an acid.
Then you make the conjugate, and you end up with excess of the weak acid, or base that you started with. That's how you make a buffer, because remember a buffer has to have two components; weak acid, and its conjugate base. Or weak base, and its conjugate acid.
So hopefully these tips, and tricks help you identify when neutralization reactions will form a buffer. Have a good one.
Please enter your name.
Are you sure you want to delete this comment?
- Tips for IDing if a Solution Is a Buffer 3,522 views
- Acid and Base Properties 25,694 views
- Strength of Acids and Bases 18,879 views
- Salts 12,012 views
- Buffered Solutions 14,204 views
- pH and pOH 20,573 views
- Bronsted-Lowry Model 12,988 views
- Conjugate Acids and Bases 13,752 views
- Equilibrium Systems 12,540 views
- Naming Acids 13,532 views
- Tips for Acid-Base Net Ionic Equations 4,266 views
- Tips for Calculating pH and pOH and More 4,019 views
- Tips for Conjugate Acid and Base Formulas 3,474 views
- Tips for Identifying Acid and Base Strength 3,802 views