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Question number 140 is an equilibrium problem involving ksp of the insoluble compound mercury 1 chloride.
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In addition, there are some unique conversions that are involved in this problem.
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The first thing they want you to do is identify the empirical formula for mercury 1 chloride.
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Mercury 1 is a diatomic ion that is hg22.
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So if hg222 plus is the cation, and chloride just has a 1 -minus charge, then we will need two chlorides for every 1 hg2 .2 .2 .2.
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Then for part b, it wants you to calculate the concentration of the hg2, 2 ,2, plus, in a saturated solution of mercury chloride.
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To do that, we need to recognize what the equilibrium is for the ksp process.
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When this dissolves, we'll get 1 mole of mercury and 2 moles of chloride, so ksp will be equal to the concentration of the mercury 1, multiplied by the chloride concentration squared.
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If the only source of chloride and mercury 2 comes from this equilibrium process, then according to the stoichiometry, the chloride concentration will be equal to two times the mercury 2 -2 -plus concentration.
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So i can then plug back into my ksp expression.
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Ksp was given to us at 1 .5 times 10 to the negative 18.
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That will then be equal to the mercury 2 -2 -plus concentration multiplied by the chloride concentration squared, which is two times the mercury 2 -2 plus concentration, and then don't forget to square it.
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So now i have one equation and one unknown, and i can solve for the mercury 2 -2 plus concentration.
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It'll be the cube root of ksp, 1 .5 times 10 to the negative 14, divided by 4, because when i square this, i'm going to square that 2 and make it 4.
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That gives me a concentration of 7 .21 times 10 to the negative 0 .4.
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Now for part c, it tells us that sea water contains 0 .2 pounds of sodium chloride per gallon.
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We can use this information to solve for the chloride concentration, and then knowing the chloride concentration and this equation up here, we can solve for the mercury concentration.
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But concentrations need to be in units of malarity in order to use the ksp expression up here.
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So we'll convert the pounds into gram.
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To do that, we recognize that for every pound there's 453 .59 grams.
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This is now the mass of sodium chloride per gallon of seawater.
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We'll then convert the gram sodium chloride into mole's sodium chloride, recognizing the molar mass of sodium chloride is 58 .44 grams per mole...