Chemistry, 22.08.2019 07:00 buddyben422

How would you prepare 1.00 l of a 0.400m solution of copper(ii)sulfate, cuso4?

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Answer from: Thelearner100

We need to dilute 0.400 mol of copper (II) sulfate, how do we know, how many weigh of CuSO_4

we have to dilute??

It's simple.

$\eta=\frac{m}{MM}$

Using a periodic table we can find the molar mass of Cu,~~S~~and~~O

$MM_{CuSO_4}=153.9~g/mol$

Then

$m=\eta*MM$

now we can replace it

m=0.400*159.6

$\boxed{\boxed{m=63.84~g}}$

Then we have to dilute 63.84 grams of copper (II) sulfate in 1 L of water to obtain a solution with 0.400M

Answer from: malissa0325

To prepare the solution of given molarity, we add 63.8 g of copper (II) sulfate to the solution.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of copper(II)sulfate solution = 0.400 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.400M=\frac{\text{Moles of copper(II)sulfate}}{1.00L}\\\\\text{Moles of copper(II)sulfate}=(0.400mol/L\times 1.00L)=0.400mol$

To calculate the mass of copper(II)sulfate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of copper(II)sulfate = 159.61 g/mol

Moles of copper(II)sulfate = 0.400 moles

Putting values in above equation, we get:

$0.400mol=\frac{\text{Mass of copper (II) sulfate}}{159.61g/mol}\\\\\text{Mass of copper (II) sulfate}=(0.400mol\times 159.61g/mol)=63.8g$

Hence, to prepare the solution of given molarity, we add 63.8 g of copper (II) sulfate to the solution.

Answer from: Quest

Sorry i dont understand the question can you rephase it?

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.