The Ksp for silver chromate (Ag2CrO4) is 1.1x10-12. Using the same initial set up of 10mL of 1M silver nitrate is combined with 25mL of 0.1M sodium chromate, what is the reaction quotient (Q) of the silver chromate? (Is a precipitate going to form?)
Ag2CrO4 ==> 2Ag^+ + CrO4^2-
Qsp = (Ag+)^2(CrO4^2-) = ?
(Ag^+) = 1 M x (10 mL/35 mL) = 0.286 M
(CrO4^2-) = 0.1 M x (25 mL/35 mL) = 0.0714 M
Plug into the above to find Qsp. If Qsp > Ksp you get a ppt.
Well, my chemistry knowledge is a little rusty, but let's give it a shot! It sounds like you're looking to calculate the reaction quotient (Q) for the formation of silver chromate.
To do that, we need to gather the necessary information:
- The initial concentrations of the reactants (silver nitrate and sodium chromate)
- The balanced chemical equation for the reaction
Now, let's dive in! Calculating Q involves multiplying the concentrations of the product and reactants, each raised to the power of their respective stoichiometric coefficients from the balanced equation.
Since you've provided the initial concentrations of silver nitrate (AgNO3) and sodium chromate (Na2CrO4), we can proceed. However, we'll need to know the balanced chemical equation that relates these two compounds to form silver chromate.
Once we have the balanced equation, we can determine the stoichiometric coefficients. From there, we can calculate the reaction quotient (Q) by plugging in the concentrations.
Now, as for whether a precipitate (solid) will form, that depends on the value of Q compared to the solubility product constant (Ksp) for silver chromate. If Q is greater than Ksp, a precipitate will form.
So, while I can't provide you with an actual answer right now, I hope this helps point you in the right direction. And remember, chemistry isn't all about formulas and equations – sometimes you just need to take a moment and appreciate the reactions.
To calculate the reaction quotient (Q) of silver chromate (Ag2CrO4), you need to determine the concentrations of the products and reactants in the reaction mixture. The reaction between silver nitrate (AgNO3) and sodium chromate (Na2CrO4) can be represented as follows:
2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3
Given:
Volume of AgNO3 solution (V1) = 10 mL = 0.01 L
Concentration of AgNO3 (C1) = 1 M
Volume of Na2CrO4 solution (V2) = 25 mL = 0.025 L
Concentration of Na2CrO4 (C2) = 0.1 M
To calculate the reaction quotient (Q), we need to determine the concentrations of Ag+ and CrO4²- ions in the reaction mixture:
[Ag+] = 2 * C1 * V1 = 2 * 1 M * 0.01 L = 0.02 M
[CrO4²-] = C2 * V2 = 0.1 M * 0.025 L = 0.0025 M
Now we can calculate the reaction quotient (Q) by multiplying the concentrations of the products raised to their stoichiometric coefficients:
Q = [Ag2CrO4] = [Ag+]² * [CrO4²-] = (0.02 M)² * (0.0025 M) = 0.000001 M³
The reaction quotient (Q) for silver chromate is 0.000001 M³.
Since the Ksp for silver chromate (Ag2CrO4) is 1.1x10^-12, and the reaction quotient (Q) is substantially smaller, it indicates that a precipitate of silver chromate should form as the reaction proceeds.
To find the reaction quotient (Q) for the formation of a precipitate, we need to write the balanced equation of the reaction and use the concentrations of the reactants.
The balanced equation is:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)
To calculate Q, we need to find the concentrations of Ag+ and CrO4^2- in the reaction mixture.
Given:
Initial volume of 1M silver nitrate (AgNO3) = 10 mL
Final volume of reaction mixture = 10 mL + 25 mL = 35 mL
Step 1: Calculate the moles of Ag+ ions
Moles of Ag+ = (initial concentration of AgNO3) * (final volume of reaction mixture)
= (1M) * (35 mL) / 1000 mL/L
= 0.035 moles
Step 2: Calculate the moles of CrO4^2- ions
Moles of CrO4^2- = (initial concentration of sodium chromate) * (final volume of reaction mixture)
= (0.1M) * (25 mL) / 1000 mL/L
= 0.0025 moles
Step 3: Calculate the concentrations of Ag+ and CrO4^2- ions
Concentration of Ag+ = (moles of Ag+) / (final volume of reaction mixture)
= 0.035 moles / 0.035 L
= 1M
Concentration of CrO4^2- = (moles of CrO4^2-) / (final volume of reaction mixture)
= 0.0025 moles / 0.035 L
= 0.0714 M
Now, we can write the expression for Q:
Q = [Ag+]^2 * [CrO4^2-]
= (1M)^2 * (0.0714 M)
= 0.0714
Comparing Q to the given Ksp value:
Since Q (0.0714) > Ksp (1.1x10^-12), a precipitate of silver chromate (Ag2CrO4) will form because the reaction quotient exceeds the solubility product constant.