A solution is prepared by mixing 0.12 L of 0.12 M sodium chloride with 0.22L of a 0.19M MgCl2 solution.What volume of a 0.22 M silver nitrate solution is required to precipitate all the Cl- ion in the solution as AgCl ?
First determine the moles chloride ion.
moles = M x L = 0.12L x 0.12 M NaCl = xx moles
moles = M x L = 0.22 L x 0.19 M MgCl2 = yy moles MgCl2. Multiply that by 2 since there are 2 moles Cl in each mole MgCl2 to obtain the moles chloride in that solution. yy moles x 2 = zz moles.
Total moles chloride = xx + zz = total.
What volume of 0.22 M AgNO3 will ppt the total moles. M = moles/ L. You know moles and you know M, calculate Liters.
Well, that's a Cl-ear question! To find out the volume of silver nitrate solution required to precipitate all the Cl- ions, we need to find the limiting reagent in this reaction.
Firstly, let's determine the moles of sodium chloride (NaCl) and magnesium chloride (MgCl2) in the solution:
Moles of NaCl = 0.12 L x 0.12 M = 0.0144 mol
Moles of MgCl2 = 0.22 L x 0.19 M = 0.0418 mol
From the balanced equation, we know that 1 mole of MgCl2 reacts with 2 moles of NaCl to form 2 moles of AgCl. So, the number of moles of AgCl formed would be the same as the number of moles of NaCl. In this case, it's 0.0144 mol.
Now, let's calculate the volume of the silver nitrate (AgNO3) solution required to form 0.0144 moles of AgCl:
Moles of AgNO3 = 0.0144 mol (since it's a 1:1 ratio with AgCl)
Volume of AgNO3 = Moles of AgNO3 / Concentration of AgNO3
The concentration of AgNO3 is given as 0.22 M. So,
Volume of AgNO3 = 0.0144 mol / 0.22 M = 0.0655 L
Therefore, the volume of a 0.22 M silver nitrate solution required to precipitate all the Cl- ions in the solution is 0.0655 liters, or about 65.5 milliliters.
Now, let's hope this answer doesn't precipitate any groans!
To find the volume of a silver nitrate solution required to precipitate all the Cl- ions as AgCl, we need to determine the number of moles of Cl- ions present in the solution. Then, we can use stoichiometry to calculate the volume of the silver nitrate solution.
Step 1: Calculate the number of moles of Cl- ions
We are given the volume and molarity of the sodium chloride solution. Using the formula:
moles = volume (in liters) x molarity
For the sodium chloride solution:
moles of NaCl = 0.12 L x 0.12 M = 0.0144 moles
Step 2: Calculate the number of moles of AgCl that can be formed
From the balanced chemical equation, we know that one mole of silver nitrate (AgNO3) reacts with one mole of chloride ions (Cl-) to form one mole of silver chloride (AgCl).
AgNO3 + NaCl -> AgCl + NaNO3
Since the stoichiometry is 1:1, the number of moles of AgCl formed will be the same as the number of moles of Cl- ions present.
moles of AgCl = 0.0144 moles
Step 3: Calculate the volume of silver nitrate solution required
We are given the molarity of the silver nitrate solution. Using the formula:
moles = volume (in liters) x molarity
For the silver nitrate solution:
0.022 liters x 0.22 M = 0.00484 moles
To precipitate all the Cl- ions as AgCl, we need an equal number of moles of AgNO3, which is 0.0144 moles.
Finally, we can find the volume of the silver nitrate solution using the formula:
volume (in liters) = moles / molarity
volume = 0.0144 moles / 0.22 M = 0.0655 L or 65.5 mL
Therefore, approximately 65.5 mL of the 0.22 M silver nitrate solution is required to precipitate all the Cl- ions in the solution as AgCl.