Calculate the volume of 0.20 mol/L KH2PO4 and the volume of 0.10 mol/L NaOH required to...

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Chemistry

Calculate the volume of 0.20 mol/L KH₂PO₄and the volume of 0.10 mol/L NaOH required to prepare 50.0 mL ofbuffer with a formal (or total) concentration ofH₂PO₄ of 0.020 mol/LÂ Â and a pH =6.3. Repeat this calculation for buffers (of the same volume andformal concentration) with pH = 6.5, 6.9, 7.5, 7.9, and 8.2.

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3.8 Ratings (428 Votes)

H2PO4^-Â Â + NaOH ----------------------------> HPO4^-2 + H2O

0.2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

0.2-0.1Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.1Â Â Â Â Â Â Â Â Â Â Â 0.1

so NaOH molarity = HPO4^- molairty

millimoles of total buffer = [H2PO4-] + [HPO4-2] = 0.02 x 50 = 1

total volume = 50

H2PO4 volume = x

HPO4^-2 = 50 -x

H2PO4- millimoles = 0.2 x

HPO4-2 millimoles = 0.1 (50 -x)

pH = pKa2 + log [salt /acid]

6.3 = 7.20 + log (0.1 (50 -x)Â Â / 0.2 x)

0.126 = 0.1 (50 -x)Â Â / 0.2 x

0.025 x = 5 - 0.1 x

x = 39.94 ml

50 -x = 10.06 ml

KH₂PO₄ volume = x = 39.94 ml

NaOH volume = 50 -x = 10.06 ml

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