Calculated the PH of the prepared buffer of 0.50 ml of 1.0M of NaOH is added...

Calculated the PH of the prepared buffer of 0.50 ml of 1.0M of NaOH is added to 50mL of the buffer.

Required Buffer (Volume, Concentration, pH)

50 mL of 0.05 M pH 12.0 buffer

Chosen Acid/ Conjugate Base Pair: Na2HPO4 / Na3PO4

Amount of Acid: 0.2498 g

Amount of Conjugate Base: 0.1227 g

Measured PH:10.92

Solutions

Expert Solution

mass of Na2HPO4 = 0.2498 g

moles of acid = 0.2498 / 141.9588 = 1.76 x 10^-3

moles of conjugate base = 0.1227 / 163.94 = 7.484 x 10^-4

moles of NaOH added = 0.5 x 1 / 1000 = 5 x 10^-4 (C)

pH = pKa + log [conjugate base + C / acid - C]

= 12.32 + log [7.484 x 10^-4 + 5 x 10^-4 / 1.76 x 10^-3 - 5 x 10^-4 ]

= 12.32

pH of the prepared buffer = 12.32

queen_honey_blossom answered 1 year ago

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