Using a 0.25 M phosphate buffer with a pH of 6.0, you add 0.79 mL of...

Using a 0.25 M phosphate buffer with a pH of 6.0, you add 0.79 mL of 0.45 M NaOH to 59 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

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Expert Solution

Since the pH = 6

It is nearer to second dissociation pKa value

pKa2 = 7.21

H2PO4 - ---------------> HPO4-2

pH = pKa2 + log [ HPO4-2] / [H2PO4-]

6 = 7.21 + log [HPO4-2] / [H2PO4-]

[HPO4-2] / [H2PO4-] = 0.061

Moles of HPO4-2 + Moles of H2PO4- = 0.25*59/1000 = 0.01475

Moles = molarity*V in L

HPO4-2 = 0.061*H2PO4-

0.061*H2PO4- + H2PO4- = 0.01475

Moles of H2PO4 - = 0.0139

MOles of HPO4-2 = 0.000848

MOles of NaOH = 0.45*0.79/1000 = 0.0003555

H2PO4- + NaOH --------> HPO4-2

0.0139 0.0003555 0.000848

0.0139-0.0003555 0 0.000848+0.0003555

pH = 7.21 + log [0.000848+0.0003555] / [0.0139-0.0003555]

pH = 6.158

queen_honey_blossom answered 1 year ago

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