# How would you determine the pH of 0.10 M NH_3 solution? NH_3 is a weak base with a K_b equal to 1.8 x 10^-5.

Jul 26, 2017

Well, we interrogate the equilibrium.......and gets $p H = 11.1 .$

#### Explanation:

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

And so ${K}_{b} = 1.8 \times {10}^{-} 5 = \frac{\left[N {H}_{4}^{+}\right] \left[H {O}^{-}\right]}{\left[N {H}_{3} \left(a q\right)\right]}$

We put in some numbers, and we assume that $x \cdot m o l \cdot {L}^{-} 1$ of ammonia ASSOCIATES, and thus at equilibrium, we know that $x = \left[N {H}_{4}^{+}\right] = \left[H {O}^{-}\right]$, and $\left[N {H}_{3}\right] = \left(0.10 - x\right) \cdot m o l \cdot {L}^{-} 1$.

And we plug in the numbers to our equilibrium expression..

${K}_{b} = 1.8 \times {10}^{-} 5 = {x}^{2} / \left(0.10 - x\right)$

And this is a quadratic in $x$, which we could solve EXACTLY if we were so-minded, but because chemists are workshy, we make the approximation that $0.10 - x \cong 0.10$. Note that we must justify this approx. later.

And thus ${x}_{1} = \sqrt{1.8 \times {10}^{-} 5 \times 0.10} = 1.34 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$.

And now that we have an approximation for $x$, we may use ${x}_{1}$ in our expression to get a second approximation........

${x}_{2} = 1.33 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

${x}_{3} = 1.33 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

This method is as exact as if we used the quadratic equation, and so.....

${x}_{3} = \left[H {O}^{-}\right] = 1.33 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$; $p O H = - {\log}_{10} \left(\left[H {O}^{-}\right]\right) = 2.88$; and $p H = 14 - p O H = 11.1 .$

We know that in aqueous solution, $14 = p O H + p H$, and so if you knows $\left[H {O}^{-}\right]$ youze also knows $\left[{H}_{3} {O}^{+}\right]$.