Consider dissolving 0.50 mol of CO2(g) to enough water to...

Consider dissolving $0.50 \mathrm{~mol}$ of $\mathrm{CO}_{2}(g)$ to enough water to make a $1.0$ -L solution. Determine the $\mathrm{pH}$ of this solution, and $\left[\mathrm{CO}_{3}{ }^{2-}\right]$. Use data from Appendix 5, Table $5.2 .$

Key Concepts

Dissolution of Gases in Water

This concept covers how a gas, once introduced into water, dissolves and may react with water to form new chemical species. In many cases, the amount of gas dissolved is dictated by its concentration and interactions in the solvent, a principle that underpins calculations of concentrations in solution.

Stepwise Acid Dissociation Equilibria

Many acids, including those formed in the carbonate system, dissociate in water in a stepwise manner. This means that the acid first loses one proton to form a conjugate base and then may lose a second proton in a subsequent equilibrium. Understanding these sequential dissociation steps is essential for determining the concentration of various species at equilibrium in a solution.

Carbonate System Equilibria

The carbonate system involves a series of equilibria among dissolved carbon dioxide, carbonic acid, bicarbonate, and carbonate ions. Recognizing how these species interconvert through acid–base reactions is crucial for predicting the pH of a solution and for calculating the relative concentrations of each component, particularly under varying conditions of pH.

pH and Equilibrium Calculations

Calculating the pH of a solution containing a weak acid or a system with multiple equilibria involves setting up equilibrium expressions based on dissociation constants and initial concentrations. This process often requires simplifying assumptions or approximations to solve for hydrogen ion concentration in a system where the pH influences the degree of dissociation and thus the distribution of species in solution.

Transcript

00:01 During the dissolving of 0 .50 moles of carbon dioxide gas in enough water to make a 1 -mole or 1 -liter solution.

00:09 We're asked to find the ph and the carbonate ion concentration.

00:15 Okay, so let's go ahead and write down our equation here.

00:25 This will produce my carbonic acid, which is a weak acid.

00:32 So this is a diprotic acid, which k -a -1 and k -a -2 are 4 .3 times 10 to the minus 7.

00:48 And 5 .6 times 10 to the minus 11.

00:55 And because ka1 is a lot bigger than ka2, the ka2 contribution will be insignificant.

01:18 So let's consider our h2o, co3, dissolution to h plus and co3 -2 -minus.

01:29 My initial concentration will be 0 .50 moles.

01:33 Per liter, zero and zero.

01:37 This will be minus x plus x plus x, making my eyes table, minus 0 .50 molar minus x, x, x and x.

01:51 My ka expression for this will be h plus times co3 divided by, and we're going to assume as follows.

02:18 We're going to say our ka, which was 4 .3 times 10 to the minus 7th, will equal x squared divided by 0 .50 minus x, which will be approximately equal to x squared over 0 .50.

02:41 And solving for x here, we will get x will equal our h plus concentration.

02:50 And let's see here.

03:18 This will equal 4 .6 times 10 to the minus 4th molar...

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