Solution Dilution With Ion Concentration | University of Arkansas - Edubirdie

University:
University of Arkansas
Course:
CHEM 1123 | University Chemistry II
Academic year:

2019
Views:

378

Pages:

1
Author:

BobN

In this example problem, we calculate the volume of stock solution needed to obtain a particular concentration of ions. Our problem reads, "You would like to make 486 milliliters of solution with a 0.5 molar nitrate ion concentration from a 1.5 molar calcium nitrate stock solution. What volume of stock solution in milliliters will you need to use?" First, we need to know the nitrate ion concentration of our stock solution since we're given the nitrate ion concentration of our target solution. Our stock solution is 1.5 molar calcium nitrate, which means it contains 1.5 moles of calcium nitrate per one liter solution. We can use the chemical formula to convert from moles of calcium nitrate to moles of just nitrate: 1 mole of calcium nitrate contains 2 moles of nitrate. This means our nitrate ion concentration in the stock solution is 3 molar nitrate. Now we can solve this problem as a solution dilution problem since we want to dilute a concentrated stock solution to create a dilute solution. We can use the equation M1V1 equals M2V2. We can plug in the dilute solution concentration and volume for M1 and V1 and the stock solution concentration for M2. If we keep the volume in milliliters for V1, then our answer for V2 will come out in milliliters. Solving for V2, we get 81 milliliters of stock solution.

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