Equilibrium constant for reaction N H 4 O H a q + dH ⊕ a q ⇌ N H 4 + a q + H 2 O l is 1.8 × 1 0 9 . Hence equilibrium constant for ionization N H 3 + H 2 O ⇌ N H 4 + a q + O H - a q x × 1 0 - 6 is.The value of ‘x’ is.

Step-by-step solution

NH 4 OH aq + H + aq ⇌ NH 4 + aq + H 2 O l K 1 = N + H 4 NH 4 OH H + = 1 .8 × 10 9 NH 4 OH aq ⇌ NH 4 + aq + OH - aq K 2 = NH 4 + OH - NH 4 OH Multiply this by H + and divide also K 2 = NH 4 + OH - H + NH 4 OH H + = K 1 × K w = 1 .8 × 10 9 × 1 .0 × 10 - 14 = 18 × 10 - 6 value of x = 18.