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2.2: Background

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    379576
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    Solubility

    We are dissolving a salt in water and like most (not all) salts the solubility increases when the temperature increases and you need to figure out the concentration of a saturated solution as a function of the temperature, make 5 measurements and plot them. In each measurement you mix known masses of the salt and the water.

    Problem 1: Figure \(\PageIndex{2}\) (a) is reading the temperature of a saturated solution, the problem is we do not know the concentration of the salt that dissolved. That is, the total salt added is the mass of the salt dissolved and the mass of the precipitate.

    \[m_{Total \; Salt} = m_{Dissolved \; Salt} \; + \; m_{Precipitated \; Salt}\]

    So even though the solution is saturated, we do no know its concentration of the solution, only the total concentration.

    three pictures, first a precipitate, second refluxing, third crystals forming on surface of glass where cold refluxed water touches hot water
    Figure \(\PageIndex{3}\): (a) left image shows a saturated solution with precipitate on the bottom, (b) middle image shows refluxing that can occur for high temperatures and (c) shows crystals forming where the cooler condensed water from the refluxing flows down the tube and hits the hotter water. (CC-BY 4.0; Robert Belford via LibreTexts)

    Problem 2: Figure \(\PageIndex{2}\) (b) shows refluxing where hot water vapor rises up inside the test tube and hits the colder glass surface that is exposed to the air and condenses. When enough condensed water forms it flows back into the solution. When this cooler water hits the bulk hot water it cools it down, figure \(\PageIndex{2}\) (c), the solubility goes down and crystals form in a ring on the surface of the test tube at the interface of the hot liquid and the air. These crystals are a false reading and due to a temperature gradient, and we do not know the temperature that they were formed at, just that it is colder than the bulk temperature, which is the temperature the thermometer is reading.

    In this experiment you will need a solution that is 28% solute by weight. This is the mass of the solute divided by the total mass of solution

    \[\%Salt=\frac{m_{salt}}{m_{salt}+m_{water}}(100)\]

     

     


    This page titled 2.2: Background is shared under a not declared license and was authored, remixed, and/or curated by Robert Belford.

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