Calculate the number of milliliters of 0.782 M KOH required to precipitate all of the Mn2+ ions in 112 mL of 0.461 M Mn(NO3)2 solution as Mn(OH)2.
To calculate the number of milliliters of 0.782 M KOH required to precipitate all of the Mn2+ ions in 112 mL of 0.461 M Mn(NO3)2 solution, we can use the following balanced chemical equation:
Mn(NO3)2 + 2KOH -> Mn(OH)2 + 2KNO3
From the equation, we can see that the ratio of Mn(NO3)2 to Mn(OH)2 is 1:1. This means that for every 1 mole of Mn(NO3)2, we need 1 mole of KOH to precipitate all the Mn2+ ions.
Step 1: Calculate the number of moles of Mn(NO3)2 in the 112 mL solution.
moles = concentration (M) x volume (L)
moles of Mn(NO3)2 = 0.461 M x (112 mL / 1000 mL/L) = 0.0516 moles
Step 2: Since the ratio of Mn(NO3)2 to Mn(OH)2 is 1:1, we need the same number of moles of KOH to react with all the Mn(NO3)2.
moles of KOH = 0.0516 moles
Step 3: Calculate the volume of 0.782 M KOH needed to provide 0.0516 moles.
volume (L) = moles / concentration (M)
volume of KOH = 0.0516 moles / 0.782 M = 0.0659 L
Step 4: Convert the volume of KOH to milliliters.
volume (mL) = 0.0659 L x 1000 mL/L = 65.9 mL
Therefore, approximately 65.9 milliliters of 0.782 M KOH are needed to precipitate all of the Mn2+ ions in 112 mL of 0.461 M Mn(NO3)2 solution as Mn(OH)2.
To calculate the number of milliliters of 0.782 M KOH required to precipitate all of the Mn²⁺ ions in the Mn(NO₃)₂ solution, we can use the stoichiometry of the balanced chemical equation and the concept of moles.
First, let's check the balanced chemical equation for the precipitation reaction between KOH and Mn(NO₃)₂:
2 KOH + Mn(NO₃)₂ → Mn(OH)₂ + 2 KNO₃
From the balanced equation, we can see that 2 moles of KOH react with 1 mole of Mn(NO₃)₂ to produce 1 mole of Mn(OH)₂.
Now, let's calculate the number of moles of Mn²⁺ ions in the 112 mL of 0.461 M Mn(NO₃)₂ solution:
moles of Mn(NO₃)₂ = concentration × volume
= 0.461 M × 0.112 L
= 0.051532 moles
Since the stoichiometry of the reaction is 1:2 between Mn(NO₃)₂ and Mn(OH)₂, the number of moles of Mn(OH)₂ required will be twice the number of moles of Mn(NO₃)₂.
moles of Mn(OH)₂ = 2 × moles of Mn(NO₃)₂
= 2 × 0.051532 moles
= 0.103064 moles
Now, we need to find out the volume (in milliliters) of 0.782 M KOH needed to react with 0.103064 moles of Mn(OH)₂.
volume of KOH (in liters) = moles of Mn(OH)₂ / concentration of KOH
= 0.103064 moles / 0.782 M
= 0.131978 L
Convert the volume to milliliters:
volume of KOH (in milliliters) = 0.131978 L × 1000
≈ 131.978 mL
Therefore, approximately 131.978 milliliters of 0.782 M KOH are required to precipitate all of the Mn²⁺ ions in 112 mL of 0.461 M Mn(NO₃)₂ solution as Mn(OH)₂.
Mn(NO3)2 + 2KOH ==> Mn(OH)2 + 2KNO3
mols Mn^2+ = M x L = ?
mols Mn(OH)2 = 2x mols Mn^2+.(look at the equation; it is 2 mols KOH for 1 mol Mn(NO3)2.
Then M KOH = mols kOH/L KOH. You know mols KOH and M KOH, solve for L KOH and convert to mL.