333.3 milliliters of 0.444 M KOH required to precipitate all of the Pb⁺² ions in 107 mL of 0.699 M Pb(NO₃)₂ solution as Pb(OH)₂.
Stoichiometry helps us use the balanced chemical equation to measure quantitative relationships and it is to calculate the amounts of products and reactants that are given in a reaction.
Molarity (M) is defined as the number of moles of solute dissolved in 1L of solution. Molarity is also known as Molar Concentration. The S.I unit of Molarity is molar (M) or mol/L.
Molarity = [tex]\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}[/tex]
Now put the values in above formula to find the number of moles of Pb(NO₃)₂
Molarity = [tex]\frac{\text{Moles of}\ Pb(NO_{3})_2}{\text{Volume of solution (in L)}}[/tex]
Moles of Pb(NO₃)₂ = Molarity × Volume
= 0.699M × 0.107 L [1ml = 0.001 L]
= 0.074 moles
The given balanced chemical equation is
Pb(NO₃)₂(aq) + 2 KOH(aq) → Pb(OH)₂(s) + 2 KNO₃(aq)
The mole ratio of KOH to Pb(NO₃)₂ is 2:1.
The mole of KOH = 0.074 mol × 2
= 0.148 moles
Now put the value in above formula to find the volume of KOH
Molarity = [tex]\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}[/tex]
Volume of KOH = [tex]\frac{\text{Number of mole of KOH}}{\text{Molarity}}[/tex]
Volume of KOH = [tex]\frac{0.148\ mol}{0.444\ M}[/tex]
= 0.333 L
= 333.3 mL [1L = 1000 mL]
Thus from the above conclusion we can say that 333.3 milliliters of 0.444 M KOH required to precipitate all of the Pb⁺² ions in 107 mL of 0.699 M Pb(NO₃)₂ solution as Pb(OH)₂.
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Question: Calculate the number of milliliters of 0.444 M KOH required to precipitate all of the Pb⁺² ions in 107 mL of 0.699 M Pb(NO₃)₂ solution as Pb(OH)₂. The equation for the reaction is:
Pb(NO₃)₂(aq) + 2 KOH(aq) → Pb(OH)₂(s) + 2 KNO₃(aq)
_________ mL KOH.
