Equilibrium constant for reaction N H 4 O H a q + H + a q ⇌ N H 4 + a q + H 2 O l is 1.8 × 19 9 . Hence equilibrium constant for ionization N H 3 + H 2 O ⇌ N H 4 + a q + O H - a q is x × 10 - 6 . The value of ' x ' is.

Step-by-step solution

N H 4 O H a q + H + a q ⇌ N H 4 + a q + H 2 O l K 1 = N H 4 + N H 4 O H H + = 1.8 × 10 9 ; N H 4 O H a q ⇌ N H 4 + a q + O H - a q K 2 = N H 4 + O H - N H 4 O H Multiply this by H + and divide also K 2 = N H 4 + O H - H + N H 4 O H H + = K 1 × K w = 1.8 × 1 0 9 × 1.0 × 1 0 - 14 = 18 × 1 0 - 6