Calculate the concentration of all ions present in each of...

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.100 mole of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ in $100.0 \mathrm{mL}$ of solution b. 2.5 moles of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ in 1.25 L of solution c. $5.00 \mathrm{g}$ of $\mathrm{NH}_{4} \mathrm{Cl}$ in $500.0 \mathrm{mL}$ of solution d. $1.00 \mathrm{g} \mathrm{K}_{3} \mathrm{PO}_{4}$ in $250.0 \mathrm{mL}$ of solution

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes.
a. 0.100 mole of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ in $100.0 \mathrm{mL}$ of solution
b. 2.5 moles of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ in 1.25 L of solution
c. $5.00 \mathrm{g}$ of $\mathrm{NH}_{4} \mathrm{Cl}$ in $500.0 \mathrm{mL}$ of solution
d. $1.00 \mathrm{g} \mathrm{K}_{3} \mathrm{PO}_{4}$ in $250.0 \mathrm{mL}$ of solution

Key Concepts

Molar Mass and Unit Conversion

Calculating the number of moles from a given mass requires knowledge of the molar mass of the substance. Additionally, converting between units of volume (e.g., milliliters to liters) is essential for accurate determination of molarity and, consequently, the concentrations of ions in solution.

Stoichiometry of Ionic Dissociation

This concept involves using the chemical formula of a compound to determine the number of ions produced upon complete dissociation. By interpreting the stoichiometric coefficients in the dissolution reaction, one can accurately calculate the concentration of each individual ion based on the molarity of the compound.

Molarity

Molarity is a concentration unit defined as the number of moles of solute per liter of solution. It serves as a fundamental measure in solution chemistry, allowing for the quantitative analysis of solute and ion concentrations when preparing or reacting solutions.

Strong Electrolytes

A strong electrolyte is a substance that completely dissociates into its constituent ions when dissolved in water. This concept is crucial because it allows one to equate the initial molarity of the compound to the sum of the concentrations of the ions produced, taking into account the stoichiometric ratios of dissociation.

Transcript

00:01 Here we are trying to solve the concentration of ions in a solution.

00:06 We're starting with this 0 .1 moles of c .a .o .3 .2 in 100 mill liters of water.

00:14 Let's take a look at the ions first, since that's what we'll be looking for in the end.

00:20 Our calcium nitrate like that in the presence of water, so basically when we put it in water, we'll just associate into calcium 2 plus and nitrate.

00:39 1 calcium, 1 calcium, 2.

00:42 We really have 1, so let's put the 2 there.

00:45 Now it's all balanced.

00:46 We can see it's a 1 to 2 ratio of calcium to nitrate.

00:54 That'll be important later.

00:57 Now let's find the total concentration.

01:00 We have 0 .1 .000 moles, and our liters is actually not going to be 100 milliliters.

01:12 It'll be 0 .10 liters, because we have to convert milliliters into a liter for this equation.

01:20 That's going to give us a molarity of one molar.

01:25 This one molar is for the whole molecule.

01:31 We can see in this one molecule we have one, calcium and two of the nitrates.

01:39 So that's our 1 to 2 ratio again.

01:43 So we have a 1 to 1 and a 1 to 2 ratio of molecule.

01:51 I'm just going to write as m to calcium.

01:53 So our calcium is still going to be 1 molar, and then we have a 1 to 2 ratio of our molecule and 2 of the nitrates for every 1 molecule.

02:06 So that's going to be 1 times 2 because 1 molar times the 2 no2s, or sorry, no3s, is going to be 2 molar.

02:19 So concentration of calcium 2 plus is 1 molar, and the concentration of the no3 minus is 2 molar.

02:35 Next we have 2 .5 moles of na2 s .4 put into 1 .25 liters of water.

02:43 Let's take a look at our ratio for every one molecule of this.

02:48 The whole molecule we have one.

02:50 Looks like we have two sodiums.

02:56 Molecule to sodium.

03:00 And looks like we only have one s .o .4 for every molecule.

03:03 So our molecule to s .4 is going to be one to one.

03:12 All right, let's solve for the moles over 1 .25 liters.

03:25 Everything's already in the correct amount it should be.

03:29 That is going to be roughly, no not rather that will be a two molar solution of the molecule.

03:44 Now for s .o4 that concentration, since it's already one to one, we don't have to multiply by anything.

03:53 It's going to be two molar because it's one molecule of s .o .4 to minus.

04:00 For the the whole molecule.

04:04 Now with the sodium, we have two sodium for every full molecule.

04:11 So that's going to be our molarity for the molecule times two.

04:15 So two times two...

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