Question

At 25 °C, how many dissociated H ions are there in 347 mL of an aqueous solution whose pH is 11.75?

Answer 1

use:
pH = -log [H+]
11.75= -log [H+]
[H+] = 1.778*10^-12 M

volume = 347 mL = 0.347 L

number of mol of H+ = CONCENTRATION * VOLUME
= (1.778*10^-12 )*0.347 L
= 6.171*10^-13 mol

number of H+ = number of mol * Avogadro's number
= (6.171*10^-13 mol)*6.022*10^23
=3.716*10^-11 H+ ions

Answer: 3.716*10^-11 H+ ions