# Tbr gen Chem buffers #8

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#### akimhaneul

##### Full Member
2+ Year Member

I don't understand how A and B leads to buffer. If you have 2 equivalents, doesn't that mean you have much more of the conjugate base or acid?

Thanks!

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#### akimhaneul

##### Full Member
2+ Year Member

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#### BerkReviewTeach

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The thing you have to consider is that a buffer is defined as a mixture of weak acid with its weak conjugate base in roughly equal concentrations, not necessarily exactly equal concentrations. If you have a 1:1 mixture, then pH = pKa and it has a large capacity to consume strong acid or strong base. If you have a mixture that is 2:1 in favor of the weak acid, then pH = pKa + log 1/2 = pKa - 0.3. If you have a mixture that is 2:1 in favor of the weak base, then pH = pKa + log 2/1 = pKa + 0.3. Both 2:1 mixtures are within the 10/1 to 1/10 range that describes a buffer.

Choice A is 1 equivalent weak acid with 2 equivalents of weak conjugate base, which forms a buffer with pH = pKa + 0.3.

Choice B is 1 equivalent weak base with 2 equivalents of weak conjugate acid, which forms a buffer with pH = pKa - 0.3.

Choice C is a diprotic acid that has been treated with 1.5 equivalents of strong base, which after reaction will leave a 1:1 mixture of the partially deprotonated form (HCO3-) and the fully deprotonated form (CO32-). This forms a buffer with pH = pKa2.

Choice D is 1 equivalent weak base with 1.5 equivalents of strong acid, which results in an over-titrated solution. After complete reaction, you have 1 equivalent of the weak conjugate acid and 0.5 equivalents of strong acid, resulting in a strong acid solution and not a buffer.

Look at Figure 9-2 on page 91 for a thorough explanation of the buffer range.

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