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# Diprotic & Polyprotic Buffers

by Jules Bruno
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So with di product buffers, remember, a die product. Acid basically has three major forms. We have the former. It has both of its acidic hydrogen. So we're gonna say that this is the acidic form. When it gives away its first acidic hydrogen, we have its intermediate form. And then when it's finally giving away its last acidic hydrogen we have it's basic form. Now, remember, if we're talking about giving away the first acidic hydrogen, that means we're going from H two a form T H a minus form A just represents non metals. We're gonna say, since we're moving that first tested a hydrogen, that means we're dealing with K one. So we're gonna say here in this example, let's say you had 0.10 Mueller Okay, carbonic acid and 0.15 Mueller of sodium bicarbonate. We can see that carbonic acid here has both of its acidic hydrogen, so it's the acidic form. This one here has one less hydrogen. So that's the conjugate base form. It only has one hydrogen left, so that's that intermediate form. In this case, we'd say that pH equals p K one. Since we're dealing with removing the first acidic hydrogen, which connects the acidic form in the intermediate form plus log of my conjugate base, which is the 0.15 Mueller sodium bicarbonate divided by the weak acid form, which is 0.10 Mueller carbonic acid. So that's the equation would use for that version. Now, if we're talking about removing the second acidic hydrogen to give us the basic form, that means we're dealing with K two. So let's say they gave us 50 m Els off 0.100 Mueller sodium bicarbonate and 30 Emmaus off 0.100 Mueller sodium carbonate. So we can see here that this form has one acidic hydrogen left and this form here has lost both acidic hydrogen. So that's the basic form in this case is were dealing with these two forms now and they're connected by K two. We'd say pH equals p k two plus log again of my conjugate base, which in this case would be the carbonate ion. Remember, we divide this by 1000 and then multiplied by the molar iti to find our moles. So they give me 10000.3 moles of my sodium carbonate divided by divide This by multiplied by the molar. It gives me 005 moles of sodium bicarbonate. So that's how we solve for the buffers, dealing with the different forms of our die product acid as we continue onward into try product buffers. It's basically the same type of idea, except we have mawr than two K values. So we're gonna have to be careful of mawr, different types of forms that exist.