How many mL of 0.109 M Na2CO3 would be needed to precipitate all of the copper ions in 26.7 mL of 0.159 M CuSO4?
To determine the amount of Na2CO3 needed to precipitate all the copper ions in CuSO4, we need to use the stoichiometry of the balanced chemical equation between Na2CO3 and CuSO4.
The balanced chemical equation is:
Na2CO3 + CuSO4 -> CuCO3 + Na2SO4
From this equation, we can see that for every 1 mole of CuSO4, we need 1 mole of Na2CO3.
First, let's calculate the number of moles of CuSO4:
Moles of CuSO4 = Volume of CuSO4 solution (in L) x Concentration of CuSO4 (in mol/L)
= 26.7 mL x (1 L / 1000 mL) x 0.159 mol/L
= 0.00425 mol
Since the stoichiometry of the reaction is 1:1 between CuSO4 and Na2CO3, we need an equal number of moles of Na2CO3 to precipitate all the copper ions.
Therefore, we need 0.00425 mol of Na2CO3.
Now let's calculate the volume of 0.109 M Na2CO3 solution required to obtain 0.00425 mol of Na2CO3:
Volume of Na2CO3 solution (in L) = Moles of Na2CO3 / Concentration of Na2CO3 (in mol/L)
= 0.00425 mol / 0.109 mol/L
= 0.0390 L
Finally, let's convert the volume to mL:
Volume of Na2CO3 solution (in mL) = Volume of Na2CO3 solution (in L) x 1000 mL/L
= 0.0390 L x 1000 mL/L
= 39.0 mL
Therefore, you would need 39.0 mL of 0.109 M Na2CO3 to precipitate all the copper ions in 26.7 mL of 0.159 M CuSO4.
To find out how many mL of the Na2CO3 solution is needed to precipitate all of the copper ions in the CuSO4 solution, we can use the concept of stoichiometry. This involves using the balanced chemical equation and the molar ratios between the reactants and products.
The balanced equation for the reaction between Na2CO3 and CuSO4 is:
Na2CO3 + CuSO4 ➞ CuCO3 + Na2SO4
From the balanced equation, we can see that the stoichiometric ratio between CuSO4 and Na2CO3 is 1:1. This means that one mole of CuSO4 requires one mole of Na2CO3 to completely react.
First, we need to calculate the number of moles of CuSO4 in the given volume of 26.7 mL:
moles of CuSO4 = volume (L) × molarity (mol/L)
= 26.7 mL × 0.159 mol/L
= 0.0042533 mol
Since the stoichiometric ratio is 1:1, the number of moles of Na2CO3 required to react with the given amount of CuSO4 is also 0.0042533 mol.
Now, we can find the volume of the 0.109 M Na2CO3 solution needed to contain this amount of moles:
volume (L) = moles ÷ molarity
= 0.0042533 mol ÷ 0.109 mol/L
= 0.039 L
To convert the volume from liters to milliliters:
volume (mL) = 0.039 L × 1000 mL/L
= 39 mL
Therefore, approximately 39 mL of the 0.109 M Na2CO3 solution would be needed to precipitate all of the copper ions in 26.7 mL of 0.159 M CuSO4.
Reaction:
Na2CO3 + CuSO4 --> CuCO3 + Na2SO4
CuCO3 is insoluble and so form ppt.
mole ratio:
Na2CO3 : CuSO4
1 : 1
actual mole of CuSO4 = Mv = 0.159M x 26.7e-3L =_______moles = moles of Na2CO3.
volume of Na2CO3 = n/M where n is the mole of Na2CO3 and M is the molarity of Na2CO3. The volume should be in L so you need to convert to mL.
hope that helps.