Question

calculate the mass of precipitate in mg that you would expect me to form when 250mL of 0.0006M Na2SO4 is added to 500mL of 0.00024 M Pb(NO3)2 (use an ice table)

Answer 1

1. Identify the given :

• Concentration of Na2SO4 = 0.0006M

Volume = 250 mL

• Concentration of Pb(NO3)2 = 0.00024M

Volume = 500mL.

• We also Know that:

m ={ w(grams) * 1000ml}/ {Mol Mass (g/mol)* V ML}

• A balanced equation of the reaction is:

Na2SO4(aq) +Pb(No3)2(aq) → PbSO4(s) + 2NaNo3(aq)

2. Calculate number of moles for Na2SO4 and Pb(NO3)2

• number of moles Na2SO4 = 0.0006M* 250ml /1000ml

=1.5x10^-4 moles (Na2SO4)

• number of moles Pb(NO3)2=0.00024M * 500mL/1000mL

= 1.2x10^-4 moles ( Pb(NO3)2)

3. Calculate the number of moles of the precipitate ( PbSO4) ; by first determining the moles proportionality from a balanced equation.

Na2SO4(aq) +Pb(No3)2(aq) → PbSO4(s) + 2NaNo3(aq)

initial : 0.00015mol +0.00012 → ------- +---------

change :0.00012 + 0.00012 → 0.00012 + 0.00024 (* by 2)

equilibri : 0.00003 + ---------- →0.00012 + 0.00024

4. Finally the mass of the precipitate (PbSO4) that formed can be calculated by :

Mass = n * Molecular mass

= 0.00012mol * 303.26 g/mol

= 0.036g

• This means that 0.036g of PbSO4 will be formed .