Today, the topic was ion concentration. We’ve already learned about the concentration of substances and molarity in this unit, so we’re going to move onto the concentration of specific ions.
As we’ve learned before, ionic compounds are made up of two parts. The first part is the cation, which is the positively charged particle. The second part is the anion, which is the negatively charged particle. When dissolved in water, these two particles separate from each other. This process is called dissociation.
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The top equation is dissociation, while the bottom is the opposite. |
When writing dissociation equations, the atoms and charges must both be balanced.
Ex.) Write the dissociation of Sodium chloride.
NaCl(s) à Na+(aq) + Cl-(aq)
Ex.) Write the dissociation for Na3PO4.
Na3PO4 à 3Na+ + PO43-
If the volume of the solution does not change, then the concentration of individual ions depends on the balanced coefficients in the dissociation equation.
Ex.) Determine [Na+] and [PO43-] in a 1.65 M solution of Na3PO4.
Na3PO4 à 3Na++ PO43-
As always, we must start with a balanced equation. Next, we convert 1.65 M to moles, and then use our molar ratios to find the number of moles of both ions. Finally, we divide by the volume. When it doesn’t tell us, we can assume it’s 1L.
1.65 M x 1 L/mol = 1.65 mol
1.65 mol x 3/1 = 4.95 mol x 1/L = 5.0 M [Na+]
1.65 mol x 1/1 = 1.65 mol x 1/L = 1.7 M [PO43-]
Done!
Ex.) A 0.100 L solution of 0.500 M PbCl2 is added to 0.200 L solution of 0.100 M NaOH. Determine the final [Cl-].
PbCl2 + NaOH à Pb+ + 2Cl- + Na+ + OH-
Just like we did before, a balanced equation is necessary. In this equation, we need to add the volumes to find the total volume. The total volume is 0.300 L. We only need to calculate the concentration of PbCl2, so we can just disregard the other parts.
0.500 M x 0.100 L/mol x 2/1 x 1/0.300L = 0.333 M [Cl-]
And that’s it! Great job!
Here’s a great video on the subject:
Posted by Michael.
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