# What is the pH of a solution whose concentration is 0.450*mol*L^-1 in ammonia; K_b=1.80xx10^-5?

Mar 6, 2017

$p H = 11.5$

#### Explanation:

We interrogate the equilibrium:

$N {H}_{3} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s N {H}_{4}^{+} + H {O}^{-}$.

We set up the equilibrium in the usual way:

${K}_{b} = 1.80 \times {10}^{-} 5 = \frac{\left[N {H}_{4}^{+}\right] \left[H {O}^{-}\right]}{\left[N {H}_{3} \left(a q\right)\right]}$

And we call the concentration of ammonia that hydrolyzes $x$.

And thus: ${K}_{b} = 1.80 \times {10}^{-} 5 = {x}^{2} / \left(0.450 - x\right)$

If, we assume that $0.450 > x$, then ${x}_{1} = \sqrt{1.80 \times {10}^{-} 5 \times 0.450}$

$= 2.85 \times {10}^{-} 3$, which is indeed small compared to $0.450$, but since we have an estimate for ${x}_{1}$, we can put it thru the washing again, and come up with:

${x}_{2} = \sqrt{1.80 \times {10}^{-} 5 \times \left(0.450 - 2.85 \times {10}^{-} 3\right)} = 2.84 \times {10}^{-} 3$

${x}_{3} = \sqrt{1.80 \times {10}^{-} 5 \times \left(0.450 - 2.84 \times {10}^{-} 3\right)} = 2.84 \times {10}^{-} 3$

Clearly I am making a meal of the successive approximations, but I think you can get the gist of the operation. Approximate, then substitute, then resubstitute, till you reach a consistent answer. Alternatively, you could have used the quadratic equation, but here there are more terms to include, and more mistakes (potentially) to make.

And thus at equilibrium,

$\left[N {H}_{4}^{+}\right] = \left[H {O}^{-}\right] = 2.84 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

$p O H = - {\log}_{10} \left[H {O}^{-}\right] = - {\log}_{10} \left(2.84 \times {10}^{-} 3\right) = 2.55$

And, since in aqueous solution, $p H + p O H = 14$, $p H = 11.45$.

If I were you, I would apply this general method to other $p H$ problems. You can get very efficient at their solutions, and you must practise how to manipulate your calculator.