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

[H₃O⁺] = 1.4 × 10⁻⁹ M.

Step-by-step 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}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* 10^(-7)\\\text{C}& -x &&&& +x &&+x\\\text{E} &1.0 - x &&&&2.5+x&&1.0* 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 (2.5\;(x + 1.0* 10^(-7)))/(1.0)`.

`\displaystyle (2.5\;(x + 1.0* 10^(-7)))/(1.0) = 1.8* 10^(-5)`.

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

Again,
`\text{K}_w = 1.0* 10^(-14)` at room temperature.

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