![]() 005 moles of sodium hydroxide, and our total volume is. So what is the resulting pH? So we're adding. 005 moles of a strong base to our buffer solution. And so our next problem is adding base to our buffer solution. So remember this number for the pH, because we're going toĬompare what happens to the pH when you add some acid and PH of our buffer solution, I should say, is equal to 9.33. And that's over theĬoncentration of our acid, that's NH four plus, and Our base is ammonia, NH three, and our concentration So the pH of our buffer solution is equal to 9.25 plus the log of the concentration So we're gonna plug that into our Henderson-Hasselbalch equation right here. Is going to give us a pKa value of 9.25 when we round. So the negative log of 5.6 times 10 to the negative 10. ![]() So let's get out the calculatorĪnd let's do that math. So the pKa is the negative log of 5.6 times 10 to the negative 10. ![]() To find the pKa, all we have to do is take the negative log of that. The Ka value for NH four plus and that's 5.6 times 10 to the negative 10. PH of our buffer solution, is to find the pKa, all right, and our acid is NH four plus. So the first thing we need to do, if we're gonna calculate the And for our problem HA, the acid, would be NH four plus and the base, A minus, would be NH three or ammonia. So we're talking about aĬonjugate acid-base pair here. Showed you how to derive the Henderson-Hasselbalch equation, and it is pH is equal to the pKa plus the log of the concentration of A minus over the concentration of HA. So now we have a strong buffer with a lot of capacity, but our pH of this buffer solution is very different from the pH of just the weak acid/conjugate base we started with.īuffer solution calculations using the Henderson-Hasselbalch equation. So, now that we're adding the conjugate base to make sure we have roughly equal amounts, our pH is no longer 2. You have to add a lot of conjugate base in order to make this be a buffer (remember, we need roughly equal amounts of the acid and its conjugate base, but in this example, the Ka is low, so we a lot of the weak acid and not much of the conjugate base). If you took a reaction of a weak acid that has a small Ka value, it will only produce some conjugate base and it's pH might be very low, like a 2. However, you have to be aware of your buffer's pH. If you have roughly equal amounts of both and relatively large amounts of both, your buffer can handle a lot of extra acid or base being added to it before being overwhelmed. It's the reason why, in order to get the best buffer possible, you want to have roughly equal amounts of the weak acid and it's conjugate base. A buffer will only be able to soak up so much before being overwhelmed. This question deals with the concepts of buffer capacity and buffer range.
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