Question

Find the \(\mathrm{pH}\) of a \(0.25 \mathrm{M}\) solution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), a strong electrolyte.

Short answer

The pH of the 0.25 M Na2CO3 solution is approximately 11.71.

Step-by-step solution

Determine the dissociation products of Na2CO3

When Na2CO3 (sodium carbonate) is dissolved in water, it dissociates into its ions: \[Na_{2}CO_{3_{(aq)}} \rightarrow 2Na^{+}_{(aq)} + CO_{3^{2-}}_{(aq)}\]

Write the chemical equations for the ionization processes

After Na2CO3 dissociates, the CO3^{2-} ions can further react with water to form HCO3^- ions and then H2CO3: \(CO_{3^{2-}}_{(aq)} + H_{2}O_{(l)} \rightleftharpoons HCO_{3^-}_{(aq)} + OH^-_{(aq)}\) (1) \(HCO_{3^-}_{(aq)} + H_{2}O_{(l)} \rightleftharpoons H_{2}CO_{3_{(aq)}} + OH^-_{(aq)}\) (2)

Calculate the concentration of all ions

Initially, the concentration of Na2CO3 is given, and we can then calculate the initial concentration of the CO3^{2-} ions: \[ [CO_{3^{2-}}]_0 = \frac{1}{2} \times [Na_{2}CO_{3}]_0 = \frac{1}{2} \times (0.25 M) = 0.125 M\] Since the concentration of hydroxide ions (OH^-) is not given, we need to calculate it using the K_b values of reactions (1) and (2), considering the K_b values of their conjugate acids in the following: \(K_{b1} = \frac{K_w}{K_{a2}}\) (For reaction 1) \(K_{b2} = \frac{K_w}{K_{a1}}\) (For reaction 2) Assuming K_w = 1.00 x 10^{-14}, K_a1 = 4.45 x 10^{-7}, and K_a2 = 4.69 x 10^{-11}: \(K_{b1} = \frac{1.00 \times 10^{-14}}{4.69 \times 10^{-11}} = 2.13 \times 10^{-4}\) \(K_{b2} = \frac{1.00 \times 10^{-14}}{4.45 \times 10^{-7}} = 2.25 \times 10^{-8}\) Now we can calculate the concentration of hydroxide ions: \[[OH^-] = \sqrt{K_{b1} [CO_{3^{2-}}]} = \sqrt{(2.13 \times 10^{-4})(0.125 M)}\] \[[OH^-] = 5.15 \times 10^{-3} M\]

Calculate the pH of the solution

Since we have the concentration of hydroxide ions, we can now find the pOH and pH of the solution: \[pOH = -\log_{10}[OH^-] = -\log_{10}(5.15 \times 10^{-3}) \approx 2.29\] Using the relationship between pH and pOH: \[pH + pOH = 14\] We can calculate the pH of the solution: \[pH = 14 - pOH = 14 - 2.29 \approx 11.71\] Thus, the pH of the 0.25 M Na2CO3 solution is approximately 11.71.

Key concepts

Dissociation of Electrolytes

Understanding the dissociation of electrolytes is crucial when calculating the pH of solutions. Electrolytes are substances that, when dissolved in water, break apart into ions, which can conduct electricity. Sodium carbonate ((Na_{2}CO_{3})) is a classic example, dissociating into sodium ions (2Na^{+}) and carbonate ions (CO_{3^{2-}}). This process is fundamental because it determines the types of ions present in the solution, which will then participate in further chemical reactions.

For students struggling with dissociation concepts, a good way to visualize this is to think of the electrolyte as a group of friends entering a party and immediately going their separate ways to mingle. Imagine each ion interacting with water much like someone mingling at a party, forming new 'conversation'—or in our context, chemical reactions.

Chemical Equilibrium

The concept of chemical equilibrium refers to the state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. Consequently, the concentrations of the reactants and products remain constant over time. It is a dynamic balance, not a static one; reactants and products are constantly being formed and broken apart at equal rates.

When we look at our sodium carbonate solution, the carbonate ion (CO_{3^{2-}}) reacts with water to form bicarbonate (HCO_{3^-}) and hydroxide ions (OH^-). The reactions reach a state of equilibrium where both the formation of HCO_{3^-} and OH^- and their recombination into CO_{3^{2-}} occur simultaneously. For students, it's like a dance floor where the number of dancers remains constant because for every person that leaves the floor, another one joins in.

Hydroxide Ion Concentration

The hydroxide ion concentration ([OH^-]) is a direct measure of the alkalinity of a solution. The more hydroxide ions present, the higher the pH, indicating a more basic (alkaline) solution. For our sodium carbonate example, after determining the concentration of carbonate ions, we can use the equilibrium constant (K_b) from related ionization reactions to calculate the hydroxide ion concentration.

It's important for students to recognize that calculating [OH^-] requires understanding both the initial concentration of the dissociated ions and the chemical equilibrium principles. A simple analogy would be like predicting the amount of air freshener needed in a room based on how many people are inside and how active they are; similarly, we predict the amount of hydroxide ions based on the initial concentration of carbonate and the activity of the equilibrium reactions.