Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically. the Fion concentration is adjusted to about 1 ppm. Some water supplies are also 'hard'; that is, they contain certain cations such as Ca2 + that interfere with the action of soap. Consider a case where the concentration of Ca2 + is 8 ppm. Could a precipitate of CaF2 form under these conditions? (Make any necessary approximations.)

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Hello everyone today we are being given the following problem says water chlorination is employed to disinfect public water supplies and swimming pools. The typical chloride ion concentration is recommended to be around one PPM or parts per million. Some water supplies contain lead that could react with the chlorine ions. If a chlorinated water sample has 0.55 parts per million of lead two plus ions with a lead chloride que sp with Eks P value of 1. times 10 to the negative fifth precipitate form. The first thing I wanna do is you want to make note that one parts per million is equal to one mg per liter which is also equal to one times 10 to the negative third grams per liter. And so for starting off With our 0.5 parts per million of lead two plus ions. This will be equal 25 times 10 to the negative 4th g/l. And then we're going to do some dimensional analysis where we take one mole is equal to 207.2 g of lead in this in form will give us our malaria. So we'll have 2.413 times 10 to the negative six malaria which is equal to moles over leaders and our units in this equation dimension analysis will cancel to give us model for leaders which will be a polarity. We will then take the fact that we have one parts per million of chloride ions which is equal to one times 10 to the negative third grams per liter. And we're going to multiply that by the Mueller mass of chlorine which is 35. g per the periodic table. When our units cancel out, we will be left with 2.82 times 10 to the negative fifth polarity. We then want to write R K. S. P expression and R K. S. P expression is going to be equal to the concentration of our products over the concentration of our reactant. We want to make note that we only include gasses and Aquarius solutions. Therefore our K. Sp expression for our lead chloride will be lead two plus in brackets. We're gonna multiply that by chloride minus and we're going to add it to their since this PB or lead chloride will associate into one lead two plus ion and two chloride ions. That coefficient becomes the exponent, hence why there's an exponent of two for the chloride. And so that case the expression for the question is gonna be able to 1.17 times 10 to the negative fifth. And so we go to calculate for Q. We'll say that Q. Is equal to 2.4131 times 10 to the negative six times our 2. times 10 to the negative fifth with the explosion of two giving us 1.92 times 10 to the negative 15. And essentially what we want to do is you want to find out if Q is greater than or equal to K and in this case Q is less than. And since Q is less than no precipitate will form, so no precipitate will form. And that is the answer for this question. Help. This helps, and until next time.

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Key Concepts

Solubility Product Constant (Ksp)

Ion Concentration and Precipitation

Hard Water and Its Effects