A concentration of 10–100 parts per billion (by mass) of Ag+ is an effective disinfectant in swimming pools. However, if the concentration exceeds this range, the Ag+ can cause adverse health effects. One way to maintain an appropriate concentration of Ag+ is to add a slightly soluble salt to the pool. Using Ksp values from Appendix D, calculate the equilibrium concentration of Ag+ in parts per billion that would exist in equilibrium with (c) AgI.

Video transcript

Hi everyone for this problem. It reads copper ions can be used to disinfect swimming pools. It is advised to use a concentration of roughly 0. parts per million to maintain this concentration, a slightly soluble salt can be added to the pool. What is the concentration of copper ions in parts per million? That is in equilibrium with copper one chloride. And were given the equilibrium constant or the scalability product? K sp at 1.7 times 10 to the negative seven. Okay, so our goal here is to calculate the concentration of copper ions. So let's go ahead and establish an equilibrium between the solid and its ions and are solid. Here is copper one chloride and we're going to establish an equilibrium with its ions. And so we have the copper ion and we have chloride ion. Okay, and K S P is our is the K S P is our solid ability product. Okay. And are solid ability. Product is going to be the product of our ions. Okay, and this is going to be our concentration of copper ions times our concentration of chlorine ions. And this value, we were given this value, it is 1.7 times 10 to the negative seven. Okay, and so in order for us to calculate the concentration of copper ions, we're going to want to calculate moller Souljah bility moller solely ability is when the concentration of our ions equal each other. Okay, so for this example let's let X represent our moller solid ability. Okay, so moller Souljah bility we'll say is X. And this is when our concentration of ions are the same. Okay, and so what we can essentially do here is solve for X. So we know what our moller soluble itty is. Ok, so essentially X is going to be squared because our concentrations are the same and we're going to set this equal to r K S P value. Okay, so we're going to have 1.7 times 10 to the negative seven is equal to X squared. And our goal here is to solve for X. X squared. So we're going to take the square root of both sides. So that means X is going to equal the square root of 1.7 times 10 to the negative seven. So when we saw for X we get a value of 4.123 times 10 to the negative four moller. And remember moller is concentration which is moles over leader. The question asks us for the concentration of copper ions in parts per million. So what we want to do now is go from malaria. T two parts per million. Okay, so alright that here. So we want to go from moles over leader two parts per million. Can also be written as a milligrams per liter. Okay, so we want to go from moles per liter, two mg per liter. So let's go ahead and do that. And we can do that using some dimensional analysis. So we'll take our value from X which is our Moeller solely ability. So we have 4.123 times 10 to the negative four moles of copper ions over one liter of solution. That's what our um elaborate E. Is. And so now we want to go from moles of copper to grams of copper and we'll do that using molar mass. So in one mole of copper Ion using our periodic table that is 63.546 g of copper. Okay, and here we can see that our moles will cancel. And we're left with grams. And remember our goal here is to go two mg per liter. So now we want to go from we have grams in our numerator. We want that to be milligrams. And so in one mg There is 10 to the -3 g. So looking at our units here are grams cancel and we're left with milligrams over leader, which was our goal. So now we have the correct units. Let's do our calculation. When we do our calculation, we get 26. parts per million. Okay, so that is going to be our concentration of copper ions and Parts per million. Or we'll just write this as 26, parts per million is our final answer. And we'll go ahead and highlight that and that is it for this problem? I hope this was helpful