Dissociation constant of water at 25°C is(a) 1.0× 10.14tion 14(b) I × 1014(d)1.8 × 10 16

Dissociation constant of water at 25°C is(a) 1.0× 10.14tion 14(b) I

1.	Dissociation constant of water at 25°C is(a) 1.0× 10.14tion 14(b) I × 1014(d)1.8 × 10 16
Answer» At 25°C, water will spontaneously dissociate very slightly into H+ and OH- ions, but the fraction is very small, only about 1 molecule in 10 million (10-7) will be dissociated. The other 9,999,999 molecules remain as molecules and don't dissociate. The reaction equation is: H2O ←→ H+ + OH- and the value of the dissociation constant (Kw means the dissociation constant for water) is defined as: Kw = [H+] [OH-] / [H2O] where the bracketed items indicate the concentration. Since only about 10-7 of the molecules dissociate, the concentrations can be expressed as [H+] = 10-7 and [OH-] = 10-7 and the value of undissociated water [H2O] is 1 -10-7 which is equal to 0.9999999 or essentially equal to 1. So Kw = (10-7)(10-7) / 1 = 10-14 That is how we get the dissociation constant of water (Kw) as 10**-14.