Question

1. Calculate the pH of a 2.5M solution of NH3 (Kb = 1.8 x 10-5 ).

Answer 1

Answer:

11.83

Explanation:

Ammonia dissolves in water according to the equilibrium expression:

$NH_{3} _{(g)} + H_{2}O_{(l)} = NH4^{+} _{(aq)} + OH^{-}_{(aq)}$

At equilibrium, a small amount dissociates, therefore:

kb = 1.8 * 10⁻⁵

pH = 2.5

the dissociation costant is given by the equilibrium amounts. This is given as:

NH₃ = (2.5 - x)

NH₄ = x

OH = x

The constant, Kb is given by:

$\frac{x^{2} }{(2.5 -x)} = 1.8 * 10^{-5}$

but x is so small that is equal to 0

Hence:

$OH^{-} = \sqrt{(1.8*10^{-5}*(2.5) }$

         = 0.006708

pOH = -log (OH)

        = -log (0.006708)

        = 2.173

pH = 14 - pOH

     = 14 - 2.173

     = 11.826

     = 11.83

Answer 2

Answer:

11.83.

Explanation:

∵ pOH = - log[OH⁻]

∵ [OH⁻] = √(Kb.C),

Kb = 1.8 x 10⁻⁵, C = 2.5 M.

∴ [OH⁻] = √(Kb.C) = √(1.8 x 10⁻⁵)(2.5 M) = 6.71 x 10⁻³ M.

∴ pOH = - log[OH⁻] = - log(6.71 x 10⁻³ M) = 2.17.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 2.17 = 11.83.