Question

At 25 °C, how many dissociated OH– ions are there in 1285 mL of an aqueous solution whose pH is 2.08?

I did pOH= 14-2.08= 11.92

[OH-] = 10^11.92 = 8.31e11

Then 8.31e10^11 *1.285 L * (6.023e23) = 6.437e35

for some reason that's not the right answer?

Answer 1

At 25 °C, how many dissociated OH– ions are there in 1285 mL of an aqueous solution whose pH is 2.08?

Answer:
pH = -log [OH^-]
but this equations gives value of H+ as moles per liter

2.5 = -log[OH^-]

[OH^-]= -Antilog 2.5 = 0.008317

[OH^-]= 0.008317 moles per liter

So, in 1285ml i.e 1.285 Liters 0.008317 ×1285/1000 = 0.010688 moles in 1285 ml

But as we know 1 mole contains Avogadro number of particles, therefore,

= 0.010688 moles in 1285 ml × 6.02 x 10²³

= 0.06434 x 10²³

Answer [OH^-] = 6.434 x 10²¹