Question

96. The dissociation constant of water at 10°C is
5.31 x 10-7. The ionic product of water is
1) 0.295 x 10-14 mole? lit-2
2) 0.6 x 10-14 mole? lit-2
3) 0.54 x 10-14 mole? lit-2
4) 0.82 x 10-14 mole? lir-2​

Answer 1

Answer:

no answer OK byy good night

Answer 2

Given info : The dissociation constant of water at 10°C is 5.31 × 10^-17 [ you made mistake in typing dissociation constant should be 5.31 × 10^-17]

To find : the ionic product of water is ...

solution : dissociation reaction of water is ...

H₂O (aq) ⇔H⁺ (aq) + OH¯ (aq)

so ionic product, k = [H⁺][OH¯]/[H₂O]...(1)

1L of water = 1000 g [ because density of water is 1g/cm³ ]

⇒no of 1 L water = 1000/18 = 55.5 mol

so, [H₂O] = 55.5 mol/L ...(2)

we also know, ionic product of water, kw = [H⁺][OH¯] ...(3)

from equation (1) , (2) and (3) we get,

k = kw/55.5

here k = 5.31 × 10^-17

⇒5.31 × 10^-17 = kw/55.5

⇒295 × 10^-17 = kw

⇒kw = 0.295 × 10¯¹⁴

Therefore the ionic product of water at 10°C is 0.295 × 10¯¹⁴