# Ammonia (NH_3) is a weak base with a K_b = 1.8 x 10^-5. What is the balanced chemical equation for the reaction of ammonia with water. Using the I.C.E the method, calculate the pH and ionization of a 1.75 M NH_3 solution in 2.50 M NH_4Cl?

Jan 24, 2017

$\textsf{p H = 9.1}$

#### Explanation:

Ammonia is a weak base and ionises in water:

$\textsf{N {H}_{3} + {H}_{2} O r i g h t \le f t h a r p \infty n s N {H}_{4}^{+} + O {H}^{-}}$

For which:

$\textsf{{K}_{b} = \frac{\left[N {H}_{4}^{+}\right] \left[O {H}^{-}\right]}{\left[N {H}_{3}\right]} = 1.8 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

These are equilibrium concentrations.

The ICE table in mol/l is:

" "sf(NH_3" "+" "H_2O" "rightleftharpoons" "NH_4^+" "+" "OH^-)

$\textsf{I \text{ "1.75" "2.5" } 0}$

$\textsf{C \text{ "-x" "+x" } + x}$

sf(E" "(1.75-x") "(2.5+x)" "x

Because the value of $\textsf{{K}_{b}}$ is sufficiently small we can assume that $\textsf{\left(1.75 - x\right) \Rightarrow 1.75}$ and $\textsf{\left(2.5 + x\right) \Rightarrow 2.5}$.

This means we can assume that the initial concentrations are a good enough approximation to those at equilibrium.

There is therefore, no need to set up an ICE table. You can just use the equation for $\textsf{{K}_{b}}$ directly and solve for $\textsf{\left[O {H}^{-}\right]}$ hence find the pH. You should always state the assumption though.

Rearranging the expression for $\textsf{{K}_{b} \Rightarrow}$

$\textsf{\left[O {H}^{-}\right] = {K}_{b} \times \frac{\left[N {H}_{3}\right]}{\left[N {H}_{4}^{+}\right]}}$

$\therefore$$\textsf{\left[O {H}^{-}\right] = 1.8 \times {10}^{- 5} \times \frac{1.75}{2.5} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{\left[O {H}^{-}\right] = 1.2 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p O H = - \log \left[O {H}^{-}\right] = - \log \left(1.2 \times {10}^{-} 5\right) = 4.92}$

$\textsf{p H + p O H = 14}$ at 298 K

$\therefore$$\textsf{p H = 14 - p O H = 14 - 4.92 = 9.08}$