# What would be the pH of a 0.503 M solution of ammonia (NH3) at room temperature? The K_b of ammonia is 1.78 x 10^-5

Apr 15, 2017

$p H \cong 11.5$

#### Explanation:

We interrogate the equilibrium...............

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

${K}_{\text{eq}} = \frac{\left[N {H}_{4}^{+}\right] \left[H {O}^{-}\right]}{\left[N {H}_{3} \left(a q\right)\right]} = 1.78 \times {10}^{-} 5$

Now we know that $\left[N {H}_{3} \left(a q\right)\right] = 0.503 \cdot m o l \cdot {L}^{-} 1$ initially, and call the amount of association it undergoes $x$.

So at equilibrium, $\left[N {H}_{4}^{+}\right] = \left[H {O}^{-}\right] = x$; and $\left[N {H}_{3} \left(a q\right)\right] = \left(0.503 - x\right) \cdot m o l \cdot {L}^{-} 1$.

${K}_{\text{eq}} = 1.78 \times {10}^{-} 5 = {x}^{2} / \left(0.503 - x\right)$

This is quadratic in $x$, which we could solve exactly, however, we make the approximation that $\left(0.503 - x\right) \cong 0.503 \cdot m o l \cdot {L}^{-} 1$.

Thus ${x}_{1} = \sqrt{1.78 \times {10}^{-} 5 \times 0.503} = 0.00299 \cdot m o l \cdot {L}^{-} 1$

Now that we an approximation for $x$, we can we substitute this value back into the equation.

And thus..........................................................

${x}_{2} = \sqrt{1.78 \times {10}^{-} 5 \times \left(0.503 - 0.00299\right)} = 0.00298 \cdot m o l \cdot {L}^{-} 1$

Since the values have converged, we can take this value as the true value.

And so $\left[H {O}^{-}\right] = \left[N {H}_{4}^{+}\right] = 2.98 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$, and thus $p O H = - {\log}_{10} \left[H {O}^{-}\right] = - {\log}_{10} \left(2.98 \times {10}^{-} 3\right) = 2.53$.

But we need $p O H$, however, we know (how?) that $p H + p O H = 14$.

And so pH=14-2.53=??