I got 4.9*10^-3 M. Thanks and
If the resulting solution has a volume of 1.9L , what is the minimum mass of CaSO4 (s) needed to achieve equilibrium?
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To find the minimum mass of CaSO4 needed to achieve equilibrium, we need to first understand the stoichiometry of the chemical reaction involved. From the given information, it seems like we have a solution of CaSO4.
The balanced chemical equation for the dissolution of CaSO4 in water is:
CaSO4(s) ⇌ Ca2+(aq) + SO42-(aq)
According to the equation, 1 mole of CaSO4 produces 1 mole of Ca2+ ions and 1 mole of SO42- ions.
To calculate the minimum mass of CaSO4 needed, we need to consider the molar concentration of Ca2+ ions in the resulting solution. We know that the concentration of Ca2+ ions is 4.9*10^-3 M.
Now, let's calculate the molar mass of Ca2+ ions using the periodic table. The molar mass of Ca2+ is 40.08 g/mol.
To convert the concentration in moles per liter (M) to moles, we need to multiply the concentration by the volume of the resulting solution:
Moles of Ca2+ = concentration of Ca2+ x volume of solution
= 4.9*10^-3 M * 1.9 L
Now we have the moles of Ca2+ ions.
To find the minimum mass of CaSO4 needed, we need to consider the stoichiometry of the reaction. From the balanced equation, we know that 1 mole of CaSO4 produces 1 mole of Ca2+ ions. Therefore, the minimum mass of CaSO4 can be calculated using the molar mass of CaSO4, which is 136.14 g/mol:
Mass of CaSO4 = moles of Ca2+ x molar mass of CaSO4
= (4.9*10^-3 M * 1.9 L) * 136.14 g/mol
By substituting the values into the equation, we can find the minimum mass of CaSO4 needed to achieve equilibrium.