Chemistry, 23.06.2019 03:30, HalpMahOnMahH0meW0rk

Calculate the ph of a .10m nh4cl solution. the kb value for nh3 is 1.8×10^-5

Answers: 1

Show answers

Answers

answered: jen4011

pH = 4.63.

Explanation

NH₄Cl is a soluble salt. It dissociates completely into its ions when dissolved in water.

NH₄Cl → NH₄⁺ + Cl⁻.

All NH₄Cl in the 0.10 M solution dissociates. There's one mole of NH₄⁺ ions in each mole formula unit of NH₄Cl. In the end, the concentration of NH₄⁺ will also be 0.10 M.

The NH₄Cl salt comes from a weak base (NH₃) and a strong acid (HCl). It would undergo hydrolysis after dissolving. Its solution will be slightly acidic.

The Kb value of NH₃ (or NH₄OH as in water) is quite small. As a result, NH₄OH tend to act as a weak base. The conjugate acid of NH₄OH is NH₄⁺. NH₄⁺ will undergo hydrolysis to produce NH₄OH and H⁺. In other words, NH₄⁺ will combine with water to produce the weak base NH₄OH and make the solution slightly acidic.

NH₄⁺ + H₂O → NH₄OH + H⁺

The initial concentration of NH₄⁺ ions is the same as that of NH₄Cl. How acidic is this solution? The acidity of this solution depends on the H⁺ concentration, which depends on the Ka value of NH₄⁺. Only the Kb of NH₃ was given. Luckily, the two species are conjugates acid and base of each other. The Kb of NH₃ relates to the Ka of NH₄⁺ by Kw, the ionization constant of water. The value of Kw under standard conditions is around 1.0 × 10⁻¹⁴.

$\text{K}_a({\text{NH}_4}^{+}) = \text{K}_w / \text{K}_b(\text{NH}_3) \\\phantom{\text{K}_a({\text{NH}_4}^{+})} = 1.0 \times 10^{-14} / (1.8 \times 10^{-5})$

Therefore, Ka(NH₄⁺) = 5.56 × 10⁻⁹.

Now make a RICE table for this reaction. Let x be the increase in the concentration of $\text{H}^{+}$ with the unit M (the same as $\text{mol} \cdot \text{L}^{-1}$ ).

$\begin{array}{cclclclc}\text{R} & {\text{NH}_4}^{+} & + & \text{H}_2\text{O} & \to & \text{NH}_4\text{OH} & + & \text{H}^{+}\\ \text{I} & 0.10 \\ \text{C} & - x & & & & +x & & + x\\ \text{E} & 0.10 -x & & & & x & & x \end{array}$

Ka of NH₄⁺ is so small that the reaction barely takes place. Note the expression for Ka.

$\text{K}_a = [\text{NH}_4\text{OH}] \cdot [\text{H}^{+}] / [{\text{NH}_4}^{+}] \\\phantom{\text{K}_a} = x^{2} / (0.10 - x)$

$x^{2} / (0.10 - x) = 5.56 \times 10^{-9}$

x will be much smaller than 0.10. (0.10 - x) will be nearly the same as 0.10. However, the x² in the numerator is a standalone term. The same approximation shall not work for x². To simplify the calculation:

$x^{2} / (0.10 - x ) \approx x^{2} / 0.10$

$x^{2} / 0.10 \approx \text{K}_a({\text{NH}_4}^{+}) = 5.56 \times 10^{-9}$

$x \approx \sqrt{5.56 \times 10^{-9} \times 0.10} = 2.36 \times 10^{-5}$ .

As a result

$[\text{H}^{+}] = x = 2.36 \times 10^{-5} \; \text{mol}\cdot \text{L}^{-1}$ ,

$\text{pH} = -\log_{10}{[\text{H}^{+}] = 4.63$ .

answered: Guest

catya your not at all which answer i've seen you do this on another question!

answered: Guest

the answer is b because, investigating whether another species of moth also drinks tears might be interesting, but it would not directly explain why this species of moth does.

to explain why this species of moth drinks tears, the scientist could ask: "what nutrients that moths need are in elephant tears? "

the scientist could experiment with moth repellent or oter ways to keep the moths away, but that would not answer his original question.