when 10.0 ml of .012 M PbNO32 is mixed with 10.0 ml of .030 M KI, a yellow precipitate of PbI2 forms.
calculate the initial molarity of Pb+2
calculate the initial molarity of I-
On measuring the equilibrium concentration of I-, it came out to be 8.0 x 10^-3 M. Calculate the molarity of I- precipitated.
To calculate the initial molarity of Pb+2, we need to consider the balanced chemical equation for the reaction:
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
Comparing the coefficients in the equation, we can see that the ratio of Pb(NO3)2 to PbI2 is 1:1. This means that the initial moles of Pb(NO3)2 and PbI2 are equal.
First, let's calculate the initial moles of Pb(NO3)2:
Moles of Pb(NO3)2 = (volume of Pb(NO3)2 solution in liters) x (molarity of Pb(NO3)2)
Given:
Volume of Pb(NO3)2 = 10.0 ml = 0.010 L
Molarity of Pb(NO3)2 = 0.012 M
Moles of Pb(NO3)2 = 0.010 L x 0.012 M = 0.00012 moles
Since the ratio of Pb(NO3)2 to PbI2 in the equation is 1:1, the initial moles of PbI2 are also 0.00012 moles.
To calculate the initial molarity of I-, we need to consider the stoichiometry of the balanced chemical equation. For every mole of Pb(NO3)2, 2 moles of KI react to form 1 mole of PbI2. Therefore, the initial moles of I- are twice the amount of moles of PbI2.
Moles of I- = 2 x moles of PbI2 = 2 x 0.00012 moles = 0.00024 moles
Now, let's calculate the initial molarity of I-:
Initial molarity of I- = (moles of I-) / (volume of KI solution in liters)
Given:
Volume of KI = 10.0 ml = 0.010 L
Molarity of KI = 0.030 M
Initial molarity of I- = 0.00024 moles / 0.010 L = 0.024 M
Now, let's calculate the molarity of I- precipitated. We know the equilibrium concentration of I- is 8.0 x 10^-3 M.
The molarity of I- precipitated can be obtained by subtracting the equilibrium concentration of I- from the initial molarity of I-.
Molarity of I- precipitated = Initial molarity of I- - Equilibrium concentration of I-
Molarity of I- precipitated = 0.024 M - 8.0 x 10^-3 M = 0.016 M