In the question Consider a 0.90 M Al(NO3)3 solution. This solution has a nitrate ion concentration of . . . .
How do you figure the nitrate ion concentration?
Al(NO3)3 ==> Al^+3 + 3NO3^-
There are three (3) nitrate ions per mol Al(NO3)3; therefore, 0.9 M Al(NO3)3 x [3 NO3^-/1 mol Al(NO3)3] = 0.9 M x 3/1 = ? M NO3^- ion
Well, to figure out the nitrate ion concentration, you need to realize that nitrates are like the life of the party in Al(NO3)3. Just imagine them wearing little party hats and dancing around. Anyway, since Al(NO3)3 is already dissolved in water, it breaks down into its ions when it's in the solution.
Now, looking at the formula, Al(NO3)3, you see that there is one aluminum ion (Al3+) and three nitrate ions (NO3-). So, for every one mole of Al(NO3)3 that dissolves, you get three moles of nitrate ions.
Since we know the concentration of the Al(NO3)3 solution is 0.90 M, it means that for every liter of the solution, you have 0.90 moles of Al(NO3)3.
So, multiplying that by three, you get 2.70 moles of nitrate ions in one liter of the solution. Therefore, the nitrate ion concentration would be 2.70 M.
Now, I hope you can picture those nitrate ions dancing away in that solution. Quite the party, huh?
To find the nitrate ion concentration in a 0.90 M Al(NO3)3 solution, you need to understand how the compound dissociates in water and calculate the molar concentration of nitrate ions (NO3-) specifically.
1. Start by identifying the chemical formula of the compound: Al(NO3)3. It consists of one aluminum ion (Al3+) and three nitrate ions (NO3-).
2. Understand the dissociation of Al(NO3)3 in water: When Al(NO3)3 is dissolved in water, it dissociates into its constituent ions according to the following equation:
Al(NO3)3 -> Al3+ + 3NO3-
This means that one molecule of Al(NO3)3 separates into one aluminum ion (Al3+) and three nitrate ions (3NO3-).
3. Determine the molar concentration of Al(NO3)3: The given solution is stated to be a 0.90 M Al(NO3)3 solution. This means that for every 1 liter of the solution, there is 0.90 moles of Al(NO3)3 present.
4. Calculate the molar concentration of nitrate ions: Since each molecule of Al(NO3)3 produces three NO3- ions, the molar concentration of nitrate ions will be three times the molar concentration of Al(NO3)3.
Nitrate ion concentration (NO3-) = 3 x Al(NO3)3 concentration
Nitrate ion concentration = 3 x 0.90 M = 2.70 M
Therefore, the nitrate ion concentration in the 0.90 M Al(NO3)3 solution is 2.70 M.