component ions. This reaction goes to completion, as indicated by the one-way arrow in the following equation: NaC2H3O2(s) ==========> Na+(aq) + C2H3O2-(aq) The Na+, being the conjugate acid of a strong base (NaOH), is too weak to react with water. However, the C2H3O2-, being the conjugate base of a weak acid (HC2H3O2), is strong enough to react slightly with water. The equation for this equilibrium reaction is: C2H3O2-(aq) + H2O(l) <———-> HC2H3O2(aq) + OH-(aq) The equilibrium constant for the above reaction (Kb for C2H3O2-) is normally not published in tables because it can be calculated from two other values that ARE published: the ionization constants for H2O and HC2H3O2. 2H2O(l) <———-> H3O+(aq) + OH-(aq) Kw = 1.0 x 10^-14 HC2H3O2(aq) + H2O(l) <———-> H3O+(aq) + C2H3O2-(aq) Ka = 1.8 x 10^-5 By combining these two equations in the appropriate way, it is possible to obtain the desired reaction and hence its equilibrium constant. What is the hydronium ion concentration ([H3O+]) in a 4.75 M NaC2H3O2 solution? answers: 3.8 x 10^-13 mol/L 4.2 x 10^-12 mol/L 1.9 x 10^-10 mol/L 5.1 x 10^-5 mol/L 2.4 x 10^-3 mol/L Can someone please explain how to do this step by step? Thank you.
Sodium acetate (NaC2H3O2) is a basic salt. When sodium acetate is dissolved in water, it dissociates into its component ions. This reaction goes to completion, as indicated by the one-way arrow in the following equation: NaC2H3O2(s) ==========> Na+(aq) + C2H3O2-(aq) The Na+, being the conjugate acid of a strong base (NaOH), is too weak to react with water. However, the C2H3O2-, being the conjugate base of a weak acid (HC2H3O2), is strong enough to react slightly with water. The equation for this equilibrium reaction is: C2H3O2-(aq) + H2O(l) <———-> HC2H3O2(aq) + OH-(aq) The equilibrium constant for the above reaction (Kb for C2H3O2-) is normally not published in tables because it can be calculated from two other values that ARE published: the ionization constants for H2O and HC2H3O2. 2H2O(l) <———-> H3O+(aq) + OH-(aq) Kw = 1.0 x 10^-14 HC2H3O2(aq) + H2O(l) <———-> H3O+(aq) + C2H3O2-(aq) Ka = 1.8 x 10^-5 By combining these two equations in the appropriate way, it is possible to obtain the desired reaction and hence its equilibrium constant. What is the hydronium ion concentration ([H3O+]) in a 4.75 M NaC2H3O2 solution? answers: 3.8 x 10^-13 mol/L 4.2 x 10^-12 mol/L 1.9 x 10^-10 mol/L 5.1 x 10^-5 mol/L 2.4 x 10^-3 mol/L Can someone please explain how to do this step by step? Thank you.
The Correct Answer and Explanation is:
Correct Answer: 1.9 × 10⁻¹⁰ mol/L
Step-by-Step Explanation:
We are asked to find the [H₃O⁺] in a 4.75 M NaC₂H₃O₂ (sodium acetate) solution. Sodium acetate is a salt of a weak acid (acetic acid) and a strong base (NaOH), so the solution is basic.
The acetate ion (C₂H₃O₂⁻) can react with water: C₂H₃O₂⁻ (aq)+H₂O (l)⇌HC₂H₃O₂ (aq)+OH⁻ (aq)\text{C₂H₃O₂⁻ (aq)} + \text{H₂O (l)} \rightleftharpoons \text{HC₂H₃O₂ (aq)} + \text{OH⁻ (aq)}
We are not given the base dissociation constant (K_b) directly, but we can calculate it: Kb=KwKa=1.0×10−141.8×10−5=5.56×10−10K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}
Now, we use an ICE table (Initial, Change, Equilibrium) for the above equilibrium reaction. Let x be the amount of OH⁻ formed:
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| C₂H₃O₂⁻ | 4.75 | -x | 4.75 – x ≈ 4.75 |
| OH⁻ | 0 | +x | x |
| HC₂H₃O₂ | 0 | +x | x |
Using the expression for Kb: Kb=x2[C2H3O2−]=x24.75K_b = \frac{x^2}{[C₂H₃O₂⁻]} = \frac{x^2}{4.75}
Substitute and solve: 5.56×10−10=x24.75⇒x2=(5.56×10−10)(4.75)=2.641×10−9⇒x=2.641×10−9≈5.14×10−55.56 \times 10^{-10} = \frac{x^2}{4.75} \Rightarrow x^2 = (5.56 \times 10^{-10})(4.75) = 2.641 \times 10^{-9} \Rightarrow x = \sqrt{2.641 \times 10^{-9}} ≈ 5.14 \times 10^{-5}
So, [OH⁻] ≈ 5.14 × 10⁻⁵ M
Now find [H₃O⁺] using: [H3O+]=Kw[OH−]=1.0×10−145.14×10−5≈1.95×10−10[H₃O⁺] = \frac{K_w}{[OH⁻]} = \frac{1.0 \times 10^{-14}}{5.14 \times 10^{-5}} ≈ 1.95 \times 10^{-10}
Rounded to two significant figures: [H3O+]=1.9×10−10 mol/L\boxed{[H₃O⁺] = 1.9 \times 10^{-10} \text{ mol/L}}
This value confirms the correct multiple-choice answer is:
1.9 × 10⁻¹⁰ mol/L.
This calculation shows how weak bases (conjugate bases of weak acids) slightly increase pH by producing OH⁻, and thus very low [H₃O⁺].
