Question

What is the [h​3​o​+​] in a solution that consists of 1.0 m nh​3​ and 2.5 m nh​4​cl? [k​b​ (nh​3​) = 1.8 ×10​-5​]a) 1.1 × 10​-5​ mb) 4.5 × 10​-5​ mc) 3.3 × 10​-9​ md) 1.4 × 10​-9​ m) 5.6 × x10​-10​ m?

Answer 1

Answer:

[H₃O⁺] = 1.4 × 10⁻⁹ M.

Explanation:

NH₄Cl is a salt that dissolves well in water. The 2.5 M NH₄Cl will give an initial NH₄⁺ concentration of 2.5 M.

NH₃ is a weak base. It combines with water to produce NH₄⁺ and OH⁻. The opposite process can also take place. NH₄⁺ combines with OH⁻ to produce NH₃ and H₂O. The final H₃O⁺ concentration can be found from the OH⁻ concentration. What will be the final OH⁻ concentration?

Let the increase in OH⁻ concentration be x. The initial OH⁻ concentration at room temperature is 10⁻⁷ M.

Construct a RICE table for the equilibrium between NH₃ and NH₄⁺:

$\begin{array}{c|ccccccc}\text{R}&\text{NH}_3 &+&\text{H}_2\text{O}&\rightleftharpoons &{\text{NH}_4}^{+}&+&\text{OH}^{-}\\\text{I}&1.0&&&&2.5&&1.0\times 10^{-7}\\\text{C}& -x &&&& +x &&+x\\\text{E} &1.0 - x &&&&2.5+x&&1.0\times 10^{-7}+x\end{array}$.

The $\text{K}_b$ value for ammonia is small. The value of x will be so small that at equilibrium, $1.0 - x \approx 1.0$ and $2.5- x \approx 2.5$.

$\displaystyle \text{K}_b = \frac{[{\text{NH}_4}^{+}]\cdot [{\text{OH}}^{-}]}{[\text{NH}_3]} \approx \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0}$.

$\displaystyle \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0} = 1.8\times 10^{-5}$.

$\displaystyle [\text{OH}^{-}] = x+1.0\times 10^{-7} = 1.8\times 10^{-5} /\left(\frac{2.5}{1.0}\right) = 7.2\times 10^{-6}\;\text{mol}\cdot\text{L}^{-}$.

Again, $\text{K}_w = 1.0\times 10^{-14}$ at room temperature.

$\displaystyle [\text{H}_3\text{O}^{+}] = \frac{\text{K}_w}{[\text{OH}^{-}]}=\frac{1.0\times 10^{-14}}{7.2\times 10^{-6}} = 1.4\times 10^{-9} \;\text{mol}\cdot\text{L}^{-1}$