Calculate Max Calcium Ion Concentration: A Chemistry Guide

by Rajiv Sharma 59 views

Hey guys! Today, we're diving into a super interesting chemistry problem: figuring out the maximum concentration of calcium ions ($Ca^{2+}$) that can chill out in a solution already hosting 0.055 M of carbonate ions ($CO_3^{2-}).We′vegotthelowdownonthesolubilityproductconstant(). We've got the lowdown on the solubility product constant (KspK_{sp})ofcalciumcarbonate() of calcium carbonate (CaCO3CaCO_3$), which is $3.4 imes 10^{-9}$. Let's break this down step by step!

Understanding the Problem

First off, let's get our heads around what's actually going on. Solubility is the name of the game here, and it's all about how much of a solid compound can dissolve in a liquid. Calcium carbonate ($CaCO_3$) isn't exactly known for its dissolving prowess in water; it's what we call a sparingly soluble salt. This means only a tiny amount of it will dissolve. The solubility product constant ($K_{sp}$), is the unsung hero here, acting as the equilibrium constant for the dissolution of a sparingly soluble salt. Think of it as the VIP pass that tells us the maximum extent to which a compound can dissolve. In simpler terms, it tells us how many ions can exist together in a solution before the compound starts precipitating out – like when it gets too crowded at a party, and some folks have to leave.

Now, when calcium carbonate decides to dissolve in water, it's not shy about showing off its ion form. It splits into calcium ions ($Ca^{2+}$) and carbonate ions ($CO_3^{2-}$). The equilibrium looks like this:

CaCO3(s)ightleftharpoonsCa2+(aq)+CO32−(aq)CaCO_3(s) ightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)

The $K_{sp}$ expression is the mathematical way we describe this equilibrium:

Ksp=[Ca2+][CO32−]K_{sp} = [Ca^{2+}][CO_3^{2-}]

Where the square brackets, like $[Ca^{2+}]$ and $[CO_3^{2-}]$, represent the molar concentrations of calcium and carbonate ions, respectively, at equilibrium. Molarity, in case you're wondering, is just a fancy way of saying how many moles of a substance are dissolved in a liter of solution (mol/L). So, the $K_{sp}$ is the product of these ion concentrations when the solution is saturated – meaning it's holding as much dissolved solute as it can without any more solid dissolving.

The Common Ion Effect

But wait, there's a twist! We already have 0.055 M of carbonate ions ($CO_3^{2-}$) floating around in our solution. This is where the common ion effect comes into play. This effect is a classic example of Le Chatelier's principle, which states that if you change the conditions of a system in equilibrium, the system will shift to counteract the change. In our case, adding carbonate ions (a common ion because it's already produced by the dissolution of $CaCO_3$) to the solution will shift the equilibrium of the dissolution reaction to the left. This means that the solubility of $CaCO_3$ will decrease, and less of it will dissolve. It's like inviting more people to an already crowded room – things get a bit cramped, and the balance shifts.

So, because we already have carbonate ions in the solution, the amount of calcium ions that can dissolve will be lower than if we were starting with pure water. This is a crucial point to keep in mind as we calculate the maximum calcium ion concentration. We're not just dealing with the inherent solubility of $CaCO_3$; we're also dealing with the push and pull of equilibrium in the presence of a common ion. This makes the problem a bit more interesting and requires us to think carefully about how the system will respond to the added carbonate ions. Now that we've got a good handle on the background, let's roll up our sleeves and get to the calculation!

Setting Up the Calculation

Okay, so we know the $K_{sp}$ for $CaCO_3$ is $3.4 imes 10^{-9}$, and the initial concentration of $CO_3^{2-}$ is 0.055 M. We need to find the maximum concentration of $Ca^{2+}$ in the solution. Let's use the $K_{sp}$ expression we talked about earlier:

Ksp=[Ca2+][CO32−]K_{sp} = [Ca^{2+}][CO_3^{2-}]

Now, let's represent the unknown concentration of $Ca^2+}$ at equilibrium as 'x'. Because $CaCO_3$ dissolves in a 11 ratio to produce $Ca^{2+$ and $CO_3^{2-}$, the increase in $Ca^{2+}$ concentration will be the same as the amount of $CaCO_3$ that dissolves. However, we already have 0.055 M of $CO_3^{2-}$ present. So, the total concentration of $CO_3^{2-}$ at equilibrium will be the initial concentration plus the amount that dissolves from $CaCO_3$, which is 0.055 + x. This is where understanding the common ion effect becomes super important.

We can now rewrite our $K_{sp}$ expression incorporating these variables:

3.4imes10−9=(x)(0.055+x)3.4 imes 10^{-9} = (x)(0.055 + x)

This equation is a quadratic equation, but we can make a simplifying assumption to make our lives easier. Since the $K_{sp}$ value is so small ($3.4 imes 10^{-9}$), we can assume that 'x' is going to be very small compared to 0.055. This is because the dissolution of $CaCO_3$ is limited, especially in the presence of the common ion ($CO_3^{2-}$). Therefore, we can approximate (0.055 + x) as simply 0.055. This simplifies our equation significantly:

3.4imes10−9\[0.2cm]approx(x)(0.055)3.4 imes 10^{-9} \[0.2cm]approx (x)(0.055)

This approximation is a game-changer because it turns a potentially messy quadratic equation into a simple linear one. It's always a good idea to check if our assumption is valid at the end, but for now, let's run with it. This simplification is based on the chemical intuition that the small $K_{sp}$ value implies very little dissolution, so the added 'x' to 0.055 is negligible. With this setup, we're ready to solve for 'x', which will give us the maximum concentration of calcium ions in the solution. Next up, we'll do the math and find our answer!

Solving for Calcium Ion Concentration

Alright, we've simplified our equation to $3.4 imes 10^{-9} \approx (x)(0.055)$. Now it's time to isolate 'x' and find the maximum calcium ion concentration. This is pretty straightforward algebra. To solve for 'x', we just need to divide both sides of the equation by 0.055:

x≈3.4imes10−90.055x \approx \frac{3.4 imes 10^{-9}}{0.055}

Let's plug those numbers into a calculator. When we do the division, we get:

x≈6.18imes10−8x \approx 6.18 imes 10^{-8}

So, the approximate maximum concentration of calcium ions ($Ca^{2+}$) in the solution is $6.18 imes 10^{-8}$ M. But before we pop the champagne, we need to check if our earlier assumption was valid. Remember, we assumed that 'x' was small enough compared to 0.055 that we could ignore it in the (0.055 + x) term. Is this a reasonable assumption?

To check, let's compare our calculated 'x' value ($6.18 imes 10^{-8}$) to 0.055. We can see that $6.18 imes 10^{-8}$ is indeed much, much smaller than 0.055. Typically, if the value we're ignoring is less than 5% of the value we're keeping, the assumption is considered valid. In this case, $6.18 imes 10^{-8}$ is so tiny compared to 0.055 that our assumption is definitely valid. If our assumption wasn't valid, we'd have to go back and solve the quadratic equation, which would be a bit more work, but in this case, we're in the clear!

Therefore, we can confidently say that the maximum concentration of calcium ions in the solution is approximately $6.18 imes 10^{-8}$ M. This result makes sense in the context of the problem. The small $K_{sp}$ value indicates that $CaCO_3$ is not very soluble, and the presence of the common ion ($CO_3^{2-}$) further reduces the solubility of $CaCO_3$, leading to a very low concentration of calcium ions in the saturated solution. Now, let's wrap things up with a final answer and a quick recap.

Final Answer and Recap

Okay, drumroll, please! The maximum concentration of calcium ions ($Ca^{2+}$) in the solution is approximately $6.18 imes 10^{-8}$ M. That's a pretty small number, which reinforces the fact that calcium carbonate isn't very soluble, especially in the presence of carbonate ions. Remember, the common ion effect played a big role here.

Let's quickly recap how we got to this answer:

  1. Understood the basics: We started by understanding the concept of solubility, the solubility product constant ($K_{sp}$), and how they relate to the dissolution of sparingly soluble salts like $CaCO_3$.
  2. Recognized the common ion effect: We identified the common ion effect, where the presence of $CO_3^{2-}$ in the solution reduces the solubility of $CaCO_3$.
  3. Set up the equilibrium expression: We wrote the $K_sp}$ expression for the dissolution of $CaCO_3$ $K_{sp = [Ca{2+}][CO_3{2-}]$.
  4. Incorporated initial conditions and changes: We accounted for the initial concentration of $CO_3^{2-}$ (0.055 M) and represented the change in concentration due to the dissolution of $CaCO_3$ with the variable 'x'.
  5. Made a simplifying assumption: We assumed that 'x' was small compared to 0.055, which allowed us to simplify the equation.
  6. Solved for 'x': We solved the simplified equation for 'x', which gave us the approximate maximum concentration of $Ca^{2+}$.
  7. Checked the assumption: We verified that our assumption was valid by comparing 'x' to 0.055.
  8. Presented the final answer: We confidently stated the maximum concentration of calcium ions as $6.18 imes 10^{-8}$ M.

So, there you have it! We successfully calculated the maximum calcium ion concentration in a solution, taking into account the common ion effect and the $K_{sp}$ value. This type of problem is a classic example of how chemical equilibrium principles can be applied to understand and predict the behavior of solutions. Keep practicing, and you'll be a solubility expert in no time! If you have any questions, drop them in the comments below. Happy chemistry-ing!