Question

use the following information to answer the next question.
ethanoic acid is a colourless liquid with the chemical formula ch₃cooh(aq). its main uses are as a chemical reactant and as a solvent.

the ph of a 0.20 mol/l solution of the above acid will be:
1.73
2.72
3.27
2.50
0.20

question 7 (1 point)
in which of the following reactions does nh₃ act as bronsted acid?
nh₃(aq) + hcl (aq) → nh₄cl (aq)
nh₃ (aq) + h⁺ (aq) → nh₄⁺ (aq)
nh₃ (aq) + na (s) → nanh₂ (aq) + ½ h₂ (g)
nh₃ (aq) + h₂o (l) → nh₄⁺ (aq) + oh⁻ (aq
nh₃(g) can never act as acid.

Explanation:

First Question (pH of Ethanoic Acid Solution)

Step1: Recall Ethanoic Acid Properties

Ethanoic acid ($\ce{CH3COOH}$) is a weak acid, so it does not dissociate completely. The acid dissociation constant ($K_a$) for ethanoic acid is approximately $1.8\times10^{-5}$.

Step2: Set Up Dissociation Equation

The dissociation of $\ce{CH3COOH}$ in water is: $\ce{CH3COOH <=> H+ + CH3COO-}$. Let $x$ be the concentration of $\ce{H+}$ ions at equilibrium. Then, at equilibrium, $[\ce{CH3COOH}] = 0.20 - x$, $[\ce{H+}] = x$, and $[\ce{CH3COO-}] = x$.

Step3: Use $K_a$ Expression

The $K_a$ expression is $K_a=\frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}$. Substituting the values, we get $1.8\times10^{-5}=\frac{x\cdot x}{0.20 - x}$. Since $K_a$ is very small, $x$ is much smaller than $0.20$, so we can approximate $0.20 - x\approx0.20$.

Step4: Solve for $x$

So, $1.8\times10^{-5}=\frac{x^2}{0.20}$. Solving for $x^2$, we get $x^2 = 1.8\times10^{-5}\times0.20 = 3.6\times10^{-6}$. Then, $x=\sqrt{3.6\times10^{-6}}\approx1.897\times10^{-3}\ \text{mol/L}$.

Step5: Calculate pH

pH is calculated as $-\log_{10}([\ce{H+}])$. So, $\text{pH}=-\log_{10}(1.897\times10^{-3})\approx2.72$.

A Bronsted acid donates a proton ($\ce{H+}$). In the reaction $\ce{NH3 (aq) + Na (s) -> NaNH2 (aq) + \frac{1}{2}H2 (g)}$, $\ce{NH3}$ donates a proton to $\ce{Na}$ (which acts as a base in this reaction), forming $\ce{NH2-}$ and $\ce{H2}$. In the other reactions, $\ce{NH3}$ is acting as a Bronsted base (accepting a proton) or the reaction is not a Bronsted - Lowry acid - base reaction in the way of $\ce{NH3}$ acting as an acid. For example, in $\ce{NH3(aq) + HCl(aq) -> NH4Cl(aq)}$, $\ce{NH3}$ accepts a proton, acting as a base. In $\ce{NH3(aq) + H+ (aq) -> NH4+ (aq)}$, $\ce{NH3}$ accepts a proton. In $\ce{NH3(aq) + H2O(l) -> NH4+ (aq) + OH- (aq)}$, $\ce{NH3}$ accepts a proton from water. And the statement that $\ce{NH3(g)}$ can never act as an acid is incorrect as shown by the reaction with $\ce{Na}$.

Answer:

2.72

Second Question (NH₃ as Bronsted Acid)