# What is pH of a solution nominally composed of 1.5*mol*L^-1 ammonia? pK_b "ammonia"=4.3xx10^-5

##### 1 Answer
Jun 16, 2017

$p H = 11.90$

#### Explanation:

We interrogate the equilibrium.......

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

for which ${K}_{b} = \frac{\left[N {H}_{4}^{+}\right] \left[H {O}^{-}\right]}{\left[N {H}_{3} \left(a q\right)\right]} = 4.3 \times {10}^{-} 5$

ANd if we propose that $x \cdot m o l \cdot {L}^{-} 1$of $\text{ammonia}$ associates, then we can rewrite the ${K}_{b}$ expression to give.......

${x}^{2} / \left(1.5 - x\right) = 4.3 \times {10}^{-} 5$, and if we make the approx., that $1.5 \text{>>} x$, then.......

$x \cong \sqrt{1.5 \times 4.3 \times {10}^{-} 5}$

${x}_{1} = 0.00803 \cdot m o l \cdot {L}^{-} 1$

${x}_{2} = 0.00801 \cdot m o l \cdot {L}^{-} 1$ (we plug the first approx. back into the expression...)

${x}_{3} = 0.00801 \cdot m o l \cdot {L}^{-} 1$. Since the approximations have converged

And thus $\left[N {H}_{4}^{+}\right] = \left[H {O}^{-}\right] = 0.00801 \cdot m o l \cdot {L}^{-} 1$, and $\left[N {H}_{3} \left(a q\right)\right] = 1.492 \cdot m o l \cdot {L}^{-} 1$.

Now $p O H = 2.10$, and thus $p H = 14 - 2.10 = 11.90$