75.0 mL 0.060 M NaF and 25 mL 0.150 M Sr (NO3) 2 are mixed. Calculate the Q.
For SrF2 Kc = 2.0 x 10-10
2NaF + Sr(NO3)2 ==>SrF2 + 2NaNO3
In the final solution before reaction:
(NaF) = 0.060 x (75/100) = ?
[Sr(NO3)2] = 0.150 x (25/100) = ?
Then Q = (Sr^2+)(F^-)^2
Plug in the final concentrations before mixing above and solve for Q.
You didn't ask but I expect it's part of the problem but
A ppt will occur IF Q > Ksp.
A ppt will not occur IF Q < Ksp
To calculate the reaction quotient (Q), we need to determine the initial concentrations of the reactants.
We have two reactants: NaF and Sr(NO3)2. The balanced chemical equation for the reaction is:
2 NaF + Sr(NO3)2 → SrF2 + 2 NaNO3
1. Calculate the initial moles of each reactant:
For NaF:
Volume (V) = 75.0 mL = 0.075 L
Concentration (C) = 0.060 M
Moles (n) = V x C = 0.075 L x 0.060 mol/L = 0.0045 mol
For Sr(NO3)2:
Volume (V) = 25.0 mL = 0.025 L
Concentration (C) = 0.150 M
Moles (n) = V x C = 0.025 L x 0.150 mol/L = 0.00375 mol
2. Calculate the reaction quotient (Q) using the moles of reactants.
Since the stoichiometric coefficients in the balanced equation are 2 for NaF and SrF2, multiply the moles of Sr(NO3)2 by 2:
Q = (mol of Sr(NO3)2 × 2) / (mol of NaF)^2
= (0.00375 mol × 2) / (0.0045 mol)^2
= 0.0075 mol / 0.00002025 mol^2
= 370.37
Therefore, the reaction quotient (Q) is 370.37.
To calculate the reaction quotient, Q, we need to find the molar concentrations of the products and reactants in the mixture.
Given:
For NaF:
Volume (V) = 75.0 mL = 0.075 L
Molarity (M) = 0.060 M
For Sr(NO3)2:
Volume (V) = 25.0 mL = 0.025 L
Molarity (M) = 0.150 M
Now, let's calculate the number of moles for each component using the formula:
moles = volume (L) x molarity (M)
For NaF:
moles of NaF = 0.075 L x 0.060 mol/L = 0.0045 mol
For Sr(NO3)2:
moles of Sr(NO3)2 = 0.025 L x 0.150 mol/L = 0.00375 mol
The balanced chemical equation for the reaction between NaF and Sr(NO3)2 is:
2NaF + Sr(NO3)2 → SrF2 + 2NaNO3
From the stoichiometry of the balanced equation, we can see that 2 moles of NaF react with 1 mole of Sr(NO3)2 to produce 1 mole of SrF2.
Therefore, the moles of SrF2 formed will be equal to half the moles of NaF used since the reaction ratio is 2:1. Thus:
moles of SrF2 = 0.0045 mol / 2 = 0.00225 mol
Now, we can calculate the concentration of SrF2:
concentration (Molarity, M) = moles / volume (L)
For SrF2:
concentration of SrF2 = 0.00225 mol / (0.075 L + 0.025 L) = 0.00225 mol / 0.10 L = 0.0225 M
Finally, we can calculate Q by substituting the concentrations of the products and reactants into the reaction quotient expression:
Q = [SrF2] / [NaF]^2
Q = 0.0225 M / (0.060 M)^2
Q ≈ 0.622
Therefore, the value of Q is approximately 0.622.