What will be the molar analytical concentration of Na CO, in the solution pro duced when 25.0 mL of 0.200 M AGNO, is mixed with 50.0 mL of 0.0800 M Na CO,?
Na2CO3 + 2AgNO3 ==> 2NaNO3 + Ag2CO3
I like to work in millimoles. If you prefer moles just divide by 1000 to find the value in mols. Working with mmols keeps all those zeros after the decimal point out of the way; like 0.05 x 0.08 = 0.004 mols or 50 x 0.08 = 4 mmols.
millimoles Na2CO3 initially = mL x M = 50 mL x 0.0800 = 4
millimoles AgNO3 initially = 25.0 x 0.200 = 5
First determine the limiting reagent (LR). Let's go through both scenarios.
a. You have 4 mmols Na2CO3. It will require 8 mmols AgNO3 and you don't have that much so AgNO3 is the LR.
b. You have 5 mmols AgNO3. It will require 2-1/2 mmols Na2CO3, You have that much Na2CO3 (and more) so AgNO3 is the LR.
Given that AgNO3 is the LR, then 4 - 2.5 = 1.5 millimols Na2CO3 remain unreacted. So (Na2CO3) = millimoles/mL = 1.5 mmols/75 mL = ?
Note that the 75 mL comes from 50 mL of Na2CO3 + 25 mL AgNO3 = 75 mL solution. Post your work if you get stuck.
Why did the sodium bicarbonate go to therapy? Because it had a "sodium COmplex" with all those molar concentrations mixed up!
To find the molar analytical concentration of NaCO3 in the solution, we need to first calculate the number of moles of NaCO3 in each solution and then add them up.
Moles of NaCO3 in 25.0 mL of 0.200 M AgNO3:
Moles = Volume (L) x Concentration (M) = (25.0 mL / 1000 mL/L) x 0.200 M = 0.005 mol
Moles of NaCO3 in 50.0 mL of 0.0800 M NaCO3:
Moles = Volume (L) x Concentration (M) = (50.0 mL / 1000 mL/L) x 0.0800 M = 0.004 mol
Adding the moles from both solutions:
Total moles of NaCO3 = 0.005 mol + 0.004 mol = 0.009 mol
To find the molar analytical concentration, we divide the total moles by the total volume of the solution in liters (75.0 mL = 0.075 L):
Molar analytical concentration of NaCO3 = Total moles / Total volume = 0.009 mol / 0.075 L = 0.12 M
So, the molar analytical concentration of NaCO3 in the solution is 0.12 M. Hope that clears things up while making you smile!
To find the molar analytical concentration of Na₂CO₃ in the solution, we need to use the concept of the dilution formula. The dilution formula is:
C₁V₁ = C₂V₂
Where:
C₁ = initial concentration of the solution (Molarity)
V₁ = initial volume of the solution (mL)
C₂ = final concentration of the solution (Molarity)
V₂ = final volume of the solution (mL)
Given:
C₁ (AGNO₃) = 0.200 M
V₁ (AGNO₃) = 25.0 mL
C₂ (Na₂CO₃) = ?
V₂ (final volume) = V₁ (AGNO₃) + V₂ (Na₂CO₃) = 25.0 mL + 50.0 mL = 75.0 mL
Now let's solve for C₂:
C₁V₁ = C₂V₂
(0.200 M)(25.0 mL) = C₂(75.0 mL)
C₂ = (0.200 M)(25.0 mL) / 75.0 mL
C₂ = 6.67 mmol/L
So, the molar analytical concentration of Na₂CO₃ in the solution is 6.67 mmol/L.
To find the molar analytical concentration of NaCO3 in the solution, we can use the concept of stoichiometry and the equation of the reaction between AgNO3 and NaCO3.
The balanced equation for the reaction is:
AgNO3 + NaCO3 → AgCO3 + NaNO3
From the equation, we can see that 1 mole of AgNO3 reacts with 1 mole of NaCO3 to produce 1 mole of AgCO3 and 1 mole of NaNO3.
First, we need to calculate the number of moles of AgNO3 and NaCO3 in the given volumes of solutions.
For AgNO3:
25.0 mL of 0.200 M AgNO3 = (25.0/1000) L × 0.200 mol/L = 0.005 mol
For NaCO3:
50.0 mL of 0.0800 M NaCO3 = (50.0/1000) L × 0.0800 mol/L = 0.004 mol
Now, we compare the coefficients in the balanced equation. The mole ratio between AgNO3 and NaCO3 is 1:1.
Since the moles of NaCO3 are less than AgNO3, NaCO3 is the limiting reagent in this reaction. This means that all of the NaCO3 will be consumed, and the amount of AgNO3 in excess will not affect the amount of NaCO3 in the final solution.
Therefore, the molar analytical concentration of NaCO3 in the solution will be equal to the initial concentration of NaCO3, which is 0.0800 M.