Question

A biochemist has 100 mL of a buffered solution at pH 7.4. The concentration of the buffer is 0.1 M, and the pKa of the buffer is 7.8. If the biochemist adds 80 mL of a 0.1 N KOH solution to the above buffered solution, what will be the final pH to the nearest tenth of a unit?

Answer 1

total millimoles of buffer = 100 x 0.1 = 10

pH = pKa + log [salt /acid]

7.4 = 7.8 + log [salt /acid]

[salt /acid] = 0.3981

[salt +acid] = 10

0.3981 acid +acid = 10

acid millimoles = 7.1525

salt millimoles = 2.8474

millimoles of KOH = 80 x 0.01 = 0.8

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

     = 7.8 + log [2.8474 + 0.8 / 7.1525 - 0.8]

     = 7.6

final pH = 7.6

Note : you gave 0.1 N KOH. check that value once . it may be 0.01 N