Question

Calculate the ph of solution obtained by mixing 100 cm3 of solution with 400cm3 of solution with ph 4

Answer 1

For pH=4, Volume = 0.1 L

[H+] = 10-4

So, mole of H+ = 10-4 * 0.1 = 10-5

For pH = 10, Volume = 0.2 L

[H+] = 10-10

As we know, Kw = 10-14 = [H+] [OH-]

So, [OH-] = 10-4

mole of OH- = 10-4 * 0.2 = 2 * 10-5

when both solution are mixed, neutralisation takes place.

So, excess mole of OH- = 2 * 10-5 - 10-5 = 10-5

Final concetration of OH- = 10-5 / (total volume)

= 10-5 / (0.1 + 0.2)

= 3.33 * 10-5

final pOH = -log[OH-] = 4.4814

So, final pH of resulting solution = 14 - pOH

= 9.518