Xcm³ of 0.25M sodium chloride was added to lead(II)nitrate until in excess,3.86g of a white precipitate were formed.
i)Write an ionic for the formation of white precipitate
ii)Work out the value of X
i) The ionic equation for the formation of the white precipitate can be written as:
2NaCl(aq) + Pb(NO3)2(aq) → 2NaNO3(aq) + PbCl2(s)
ii) To calculate the value of X, we need to determine the number of moles of the white precipitate formed.
First, we need to calculate the number of moles of the white precipitate (PbCl2). The molar mass of PbCl2 is 278.1 g/mol (from the periodic table).
Using the equation:
mass = moles × molar mass,
we can calculate the moles of PbCl2:
3.86 g = moles × 278.1 g/mol
Solving for moles, we get:
moles = 3.86 g / 278.1 g/mol = 0.0139 mol
Since 1 mole of PbCl2 corresponds to 2 moles of NaCl, we can calculate the moles of NaCl added using the mole ratio:
moles of NaCl = 0.0139 mol × (2 mol NaCl / 1 mol PbCl2) = 0.0277 mol
Finally, we can calculate the volume (X) using the molarity of the NaCl solution:
Molarity = moles / volume
0.25 M = 0.0277 mol / X
Solving for X, we get:
X = 0.0277 mol / 0.25 M = 0.1108 L = 110.8 cm³
Therefore, the value of X is 110.8 cm³.