At 25 °C, how many
dissociated OH− ions are there
in 1291 mL of an aqueous solution whose pH
is 1.35?
use:
pH = -log [H+]
1.35 = -log [H+]
[H+] = 4.467*10^-2 M
use:
[OH-] = Kw/[H+]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[OH-] = (1.0*10^-14)/[H+]
[OH-] = (1.0*10^-14)/(4.467*10^-2)
[OH-] = 2.239*10^-13 M
We have:
volume , V = 1.291*10^3 mL
= 1.291 L
use:
number of mol,
n = Molarity * Volume
= 2.239*10^-13*1.291
= 2.891*10^-13 mol
use:
number of OH- ions = number of moles * Avogadro’s number
= 2.891*10^-13 * 6.022*10^23 OH- ions
= 1.741*10^11 OH- ions
Answer: 1.741*10^11 OH- ions
At 25 °C, how many dissociated OH− ions are there in 1291 mL of an aqueous solution whose pH is 1.35?
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