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A buffer consists of 0.24 M NaH2PO4 and 0.47 M Na2HPO4. Given that the K values...

A buffer consists of 0.24 M NaH2PO4 and 0.47 M Na2HPO4. Given that the K values for H3PO4 are, Ka1 = 7.2 x 10-3, Ka2 = 6.3 x 10-8, and Ka3 = 4.2 x 10-13, calculate the pH for this buffer.

I know that the answer is 7.49 because pH = -log(6.3 x 10-8) + log(0.47/0.24). What I want to know is WHY 6.3 x 10-8 was used instead of Ka1 or Ka3. Thanks.

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Answer #1

liven [NaH₂Pou] = 0.24M CNCY HPO4) = 0.47M CH2POU ] = 0:2UM CHPOU?) = 0:47M solution is butter of mixtere of weak wid Hapoy a

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