Solution contains [Cu 2+ ] = 0.035 M, [Sr 2+ ] = 0.054 M, [Al 3+ ] = 0.23 M. Cu 2+ can be separated by selective precipitation using NaOH. What is the minimum concentration of NaOH needed to start precipitation of Cu 2+ ? (K sp = 2.2 × 10 − 20 of Cu(OH) 2 ).

Your tool that a solution contains copper 2 ion concentration of 0.035 molar, strontium ion concentration of 0.05 4 molar, and aluminum ion concentration of 0.23 molar. Copper 2 can be separated by selective precipitation using sodium hydroxide. What is the minimum concentration of sodium hydroxide needed to precipitate start precipitation of copper 2 ion? Here ksp equals 2.2 times 10 to the negative 20 of copper 2 hydroxide. Alright, so they're giving us ksp of copper 2 hydroxide, so this is our point of interest. We're going to say since they're giving us Ksp of that, that is our ionic solid, and we show how it breaks up into its ions. It breaks up into copper 2 ion, plus 2 hydroxide ions. And since we're dealing with KSP, we're dealing with an ICE chart, so initial change equilibrium. Remember, in an ICE chart we ignore solids and liquids. Now the initial concentration of copper is this value, 0.035 molar. We're not given an initial amount of hydroxide ion, so it's 0. They're both products who are making them, so this is plus x plus there's a 2 year 2 x, bring down everything. So we've filled out our ICE chart. Now here we're going to say that k s p equals products, so it equals copper 2 ion times hydroxide ion squared. So here Ksp is 2.2 times 10 to the negative 20 equals now when you have a real number before your variable of x, you can ignore your variable of x. That's because x is so small, it's not gonna have a great enough impact on that real number. So here the concentration would be 0.035. Hydroxide is 2 x still, and it's squared, so that's 0.035 times 4 x squared, still being equal to my ksp. Now do 0.035 times 4, so that's 2.20 times 10 to the negative 20 equals 0.14 x squared. Again, that's 0.035 times 4 gives me that 0.14. Divide both sides by 0.14, So x squared equals 1.571 times 10 to the negative 19. Take the square root of both sides and that'll give me my x. X here equals 3.964 times 10 to the negative 10. Here though, they were asking for the minimum concentration of hydroxide ion. We found x, and we're gonna say that x, we have to relate it back to sodium hydroxide. In our equation, hydroxide is part of sodium hydroxide and at equilibrium it equals 2x. So here we're going to say the concentration of o h minus equals 2x. So plug in what we just found for x, so plug it in here. So the concentration of hydroxide ion equals 7.93 times 10 to the negative 10 molar. Now remember that the concentration of hydroxide ions would be equal to the concentration of our strong base because they break up in a one to 1 ratio. So here, by finding the concentration of hydroxide ions, we are finding the concentration of sodium hydroxide. So again, our final answer would be 7.93 times 10 to the negative 10 molar.

Solution contains [Cu2+] = 0.035 M, [Sr2+] = 0.054 M, [Al3+] = 0....

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Solution contains [Cu²⁺] = 0.035 M, [Sr²⁺] = 0.054 M, [Al³⁺] = 0.23 M. Cu²⁺ can be separated by selective precipitation using NaOH. What is the minimum concentration of NaOH needed to start precipitation of Cu²⁺? (K_sp = 2.2 × 10⁻²⁰ of Cu(OH)₂).

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Solution contains [Cu2+] = 0.035 M, [Sr2+] = 0.054 M, [Al3+] = 0.23 M. Cu2+ can be separated by selective precipitation using NaOH. What is the minimum concentration of NaOH needed to start precipitation of Cu2+? (Ksp = 2.2 × 10−20 of Cu(OH)2).

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Selective Precipitation practice set

Solution contains [Cu²⁺] = 0.035 M, [Sr²⁺] = 0.054 M, [Al³⁺] = 0.23 M. Cu²⁺ can be separated by selective precipitation using NaOH. What is the minimum concentration of NaOH needed to start precipitation of Cu²⁺? (K_sp = 2.2 × 10⁻²⁰ of Cu(OH)₂).