How ml silver nitrate solution (0.05 mol / l) K = 1.0000 consumed for titration of sodium chloride, 0.0500 g. MW Sodium chloride 58.44.
Ag^+ + Cl^- ==> AgCl
mols NaCl = 0.05g/molar mass = ?
mols AgNO3 = mols NaCl
M AgNO3 = mols AgNO3/L AgNO3
You know M and mols, solve for L AgNO3 and convert to mL.
To determine the amount of silver nitrate solution required for the titration of sodium chloride, we need to use stoichiometry and the given information.
First, let's convert the given mass of sodium chloride (NaCl) to moles:
moles of NaCl = mass / molar mass
moles of NaCl = 0.0500 g / 58.44 g/mol = 0.0008563 mol
Since the balanced chemical equation between silver nitrate (AgNO3) and sodium chloride (NaCl) is 1:1, the moles of silver nitrate required will also be 0.0008563 mol.
Now, we need to determine the volume (in liters) of the silver nitrate solution needed. We can use the molarity (0.05 mol/L) of the silver nitrate solution to find this.
moles = volume (in liters) x molarity
0.0008563 mol = volume (in liters) x 0.05 mol/L
Rearranging the equation, we find:
volume (in liters) = moles / molarity
volume (in liters) = 0.0008563 mol / 0.05 mol/L = 0.017125 L
Therefore, to titrate 0.0500 g of sodium chloride using a silver nitrate solution with a concentration of 0.05 mol/L, you would need approximately 0.017125 liters (or 17.125 mL) of the silver nitrate solution.