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Using a 0.30 M phosphate buffer with a pH of 6.3, you add 0.80 mL of...

Using a 0.30 M phosphate buffer with a pH of 6.3, you add 0.80 mL of 0.50 M NaOH to 60. mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

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Answer #1

millimoles of buffer = 0.30 x 60 = 18

H2PO4-   ----------------------> HPO42- + H+

pKa2 = 7.21

pH = 6.3

pH = pKa + log [HPO42- / H2PO4-]

6.3 = 7.21 + log [HPO42- / H2PO4-]

[HPO42- / H2PO4-] = 0.123

HPO42- = 0.123 H2PO4-

HPO42- + H2PO4- = 18

0.123 H2PO4- + H2PO4- = 18

H2PO4- = 16.02

HPO42- = 1.97

millimoles of NaOH    (C) = 0.80 x 0.5 = 0.4

pH = pKa + log [salt + C / acid - C]

      = 7.21 + log [1.97 + 0.4 / 16.02 - 0.4]

       = 6.39

pH = 6.39

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