Chromate Ion Mass: A Chemistry Calculation Guide
Hey chemistry whizzes! Today, we're diving deep into a classic problem that often pops up in aqueous solutions: calculating the mass of specific ions. Specifically, we'll be tackling how to figure out the mass of chromate ions (CrO₄²⁻) when calcium chromate (CaCrO₄) is introduced into water. This might sound a bit technical, but trust me, guys, once you break it down, it's totally manageable and a fantastic way to sharpen those chemical calculation skills. We're aiming to find the mass in milligrams, rounded to the nearest whole number, after dissolving CaCrO₄ in 100 cm³ of water, assuming the final volume remains 100 cm³. So, grab your calculators, a notebook, and let's get this done!
Understanding Solubility and Saturated Solutions
Before we can even think about calculating the mass of chromate ions, we need to get a handle on what a saturated solution actually is. Imagine you're adding sugar to your tea. You keep stirring, and the sugar dissolves. Eventually, you reach a point where no matter how much more sugar you add, it just sinks to the bottom. That's your saturated solution! In chemistry terms, a saturated solution is a solution in which the maximum amount of solute has been dissolved at a given temperature and pressure. Any additional solute added will not dissolve and will remain as a solid precipitate. For our CaCrO₄ problem, this means we need to know its solubility product constant (Ksp). The Ksp is a crucial value that tells us how soluble a sparingly soluble salt is. A lower Ksp value indicates lower solubility, meaning less of the salt will dissolve to form ions in the solution. The Ksp expression is derived from the dissolution equilibrium equation. For CaCrO₄, the dissolution process is:
CaCrO₄(s) <=> Ca²⁺(aq) + CrO₄²⁻(aq)
And its Ksp expression is:
Ksp = [Ca²⁺][CrO₄²⁻]
Here, [Ca²⁺] and [CrO₄²⁻] represent the molar concentrations of calcium and chromate ions, respectively, in a saturated solution. The critical part here is that for every mole of CaCrO₄ that dissolves, you get one mole of Ca²⁺ ions and one mole of CrO₄²⁻ ions. This 1:1 stoichiometry is super important for our calculations. So, to figure out how much CaCrO₄ dissolves, and consequently, how many chromate ions are present, we absolutely need the Ksp value for CaCrO₄. Without it, we're flying blind! Make sure you always look up the correct Ksp for the specific compound you're working with, as these values can vary slightly depending on the source, but they are generally accepted within a certain range. Remember, the Ksp is temperature-dependent, so if a temperature isn't specified, we usually assume room temperature (around 25°C), which is when most standard Ksp values are reported. This understanding of solubility and Ksp is the bedrock upon which all our subsequent calculations will be built. It's the key to unlocking the concentration of ions in our solution.
The Solubility Product Constant (Ksp) for Calcium Chromate
Okay, so we've established that the solubility product constant (Ksp) is our guiding star for this calculation. For calcium chromate (CaCrO₄), this constant is essential because CaCrO₄ is considered a sparingly soluble salt. This means it doesn't dissolve completely in water, and we're dealing with an equilibrium between the solid salt and its dissolved ions. The Ksp value essentially quantifies this equilibrium. For CaCrO₄, the dissolution reaction is:
CaCrO₄(s) <=> Ca²⁺(aq) + CrO₄²⁻(aq)
The Ksp expression is then written as:
Ksp = [Ca²⁺][CrO₄²⁻]
Now, what is this value? A commonly accepted Ksp value for CaCrO₄ at 25°C is approximately 7.1 x 10⁻³³. Yes, that's a tiny number, which confirms its sparingly soluble nature. This low Ksp tells us that the concentration of dissolved ions in a saturated solution will be very, very small. When we introduce solid CaCrO₄ into water, it will dissolve until the product of the calcium ion concentration and the chromate ion concentration reaches this Ksp value. Because the stoichiometry of the dissolution is 1:1 (one mole of CaCrO₄ yields one mole of Ca²⁺ and one mole of CrO₄²⁻), the molar concentration of Ca²⁺ ions will be equal to the molar concentration of CrO₄²⁻ ions in the saturated solution, assuming no other sources of these ions are present. Let's denote the molar solubility of CaCrO₄ as 's'. Then, at equilibrium, [Ca²⁺] = s and [CrO₄²⁻] = s. Substituting these into the Ksp expression, we get:
Ksp = (s)(s) = s²
Therefore, s = √Ksp. This 's' represents the molar concentration of both Ca²⁺ and CrO₄²⁻ ions in the saturated solution. Understanding this relationship is key. We're not just dissolving a fixed amount; we're reaching an equilibrium where the ion concentrations are dictated by the Ksp. The Ksp value acts as a ceiling for the product of ion concentrations. It's a fundamental constant that governs the maximum solubility of the salt in water. So, whenever you're dealing with solubility calculations for ionic compounds, identifying and using the correct Ksp is your first and most critical step. It's the language the salt speaks when it tries to dissolve.
Calculating Molar Solubility
Alright guys, now that we've got our Ksp for CaCrO₄ (which is 7.1 x 10⁻³³), we can move on to calculating the molar solubility of the salt. Molar solubility, often represented by the letter 's', is the concentration of the dissolved salt in moles per liter (mol/L) in a saturated solution. As we established earlier, when CaCrO₄ dissolves, it dissociates into one calcium ion (Ca²⁺) and one chromate ion (CrO₄²⁻):
CaCrO₄(s) <=> Ca²⁺(aq) + CrO₄²⁻(aq)
Because the stoichiometry is 1:1, for every mole of CaCrO₄ that dissolves, we get one mole of Ca²⁺ and one mole of CrO₄²⁻. This means that in a saturated solution, the molar concentration of calcium ions, [Ca²⁺], will be equal to 's', and the molar concentration of chromate ions, [CrO₄²⁻], will also be equal to 's'.
We use the Ksp expression we discussed:
Ksp = [Ca²⁺][CrO₄²⁻]
Substituting 's' for both ion concentrations, we get:
Ksp = (s)(s) = s²
Now, we can solve for 's' by taking the square root of the Ksp value:
s = √Ksp
Plugging in our Ksp value for CaCrO₄:
s = √(7.1 x 10⁻³³)
Let's crunch those numbers. The square root of 7.1 x 10⁻³³ is approximately 2.66 x 10⁻¹⁷.
So, the molar solubility 's' of CaCrO₄ is 2.66 x 10⁻¹⁷ mol/L. This means that in a saturated solution of calcium chromate, the concentration of Ca²⁺ ions is 2.66 x 10⁻¹⁷ mol/L, and the concentration of CrO₄²⁻ ions is also 2.66 x 10⁻¹⁷ mol/L. This is an incredibly low concentration, which is consistent with the very small Ksp value. This molar solubility is the key to finding out how much of our target ion, the chromate ion, is actually dissolved and present in the solution. We've successfully determined the concentration of chromate ions in moles per liter, which is a huge step forward!
Converting Moles to Mass
We've calculated the molar solubility of CaCrO₄ and found that the concentration of chromate ions ([CrO₄²⁻]) in our saturated solution is 2.66 x 10⁻¹⁷ mol/L. Now, the question asks for the mass of chromate ions in milligrams. To get there, we need to convert moles to mass. This involves using the molar mass of the chromate ion (CrO₄²⁻).
First, let's find the molar mass of the chromate ion. The formula is CrO₄²⁻. We need the atomic masses of Chromium (Cr) and Oxygen (O). From the periodic table:
- Atomic mass of Cr ≈ 51.996 g/mol
- Atomic mass of O ≈ 15.999 g/mol
The molar mass of CrO₄²⁻ is calculated as:
Molar mass (CrO₄²⁻) = (1 x Atomic mass of Cr) + (4 x Atomic mass of O)
Molar mass (CrO₄²⁻) = (1 x 51.996 g/mol) + (4 x 15.999 g/mol)
Molar mass (CrO₄²⁻) = 51.996 g/mol + 63.996 g/mol
Molar mass (CrO₄²⁻) ≈ 115.992 g/mol
So, one mole of chromate ions has a mass of approximately 115.992 grams. Now we can convert the molar concentration of chromate ions to a mass concentration. We know the concentration is 2.66 x 10⁻¹⁷ mol/L. To find the mass of chromate ions in one liter of solution, we multiply the molar concentration by the molar mass:
Mass concentration (g/L) = Molar concentration (mol/L) x Molar mass (g/mol)
Mass concentration (g/L) = (2.66 x 10⁻¹⁷ mol/L) x (115.992 g/mol)
Mass concentration (g/L) ≈ 3.085 x 10⁻¹⁵ g/L
This means that in one liter of saturated CaCrO₄ solution, there are approximately 3.085 x 10⁻¹⁵ grams of chromate ions. Pretty wild, right? That's an incredibly tiny amount of mass. The final step is to convert this mass from grams to milligrams, as requested by the problem.
- 1 gram (g) = 1000 milligrams (mg)
So, we multiply our mass concentration in g/L by 1000:
Mass concentration (mg/L) = (3.085 x 10⁻¹⁵ g/L) x (1000 mg/g)
Mass concentration (mg/L) ≈ 3.085 x 10⁻¹² mg/L
This is the mass of chromate ions per liter of saturated solution. We're almost there, guys!
Final Calculation for 100 cm³ Solution
We've calculated the mass concentration of chromate ions in a saturated solution to be approximately 3.085 x 10⁻¹² mg/L. The problem states that we have 100 cm³ of solution and that we should assume the final volume is 100 cm³. The first thing to do is convert the volume from cubic centimeters (cm³) to liters (L), because our concentration is in mg/L.
- 1 Liter (L) = 1000 cubic centimeters (cm³)
So, 100 cm³ is equal to:
Volume (L) = 100 cm³ / 1000 cm³/L = 0.1 L
Now, we can calculate the total mass of chromate ions in this 0.1 L of solution by multiplying the volume by the mass concentration:
Total Mass (mg) = Volume (L) x Mass concentration (mg/L)
Total Mass (mg) = (0.1 L) x (3.085 x 10⁻¹² mg/L)
Total Mass (mg) = 0.3085 x 10⁻¹² mg
This can also be written as 3.085 x 10⁻¹³ mg.
The problem asks for the result in milligrams, rounded to the nearest whole number. Looking at our result, 3.085 x 10⁻¹³ mg, this number is extremely small. When we round this to the nearest whole number, it becomes 0 mg.
So, the mass of chromate ions in 100 cm³ of a saturated CaCrO₄ solution, given the assumptions, is 0 milligrams when rounded to the nearest whole number. This highlights just how sparingly soluble calcium chromate is. Even though there are ions present, their total mass in such a small volume is far below the precision of a whole milligram. It’s a great illustration of how Ksp values dictate very low concentrations for poorly soluble salts.
Final Answer: The mass of chromate ions in the saturated solution is 0 mg (rounded to the nearest whole number).