Question

3 Pb(NO3)2 + 2Na3PO4----> Pb3(PO4)2 + 6 NaNO3

If 50.0 ml of both solutions were mixed together and 1.41 g of solid presipitate was obrained, what would be the molarity of the resulting NaNO3 solution? (Assume the volume of the prescipitate is negligible compared to the volume of the solution and that the volumes are additive)

3KOH + H3Z---> K3Z + 3H2O H3Z-> TRIPROTIC ACID

If a 30 ml of 0.0250 M solution H3Z is titrated with a 0.105 M of the base, how many mL of the base will be required to beautralize all of the acid?

Answer 1

a)

Chemical Equation

3Pb(NO3)2 (aq) + 2Na3PO4 (aq) -------> Pb3(PO4)2 (s) + 6 NaNO3 (aq)

Mass of Solid Presipitate (  Pb3(PO4)2 ) = 1.41 g

Number of moles of  Pb3(PO4)2 = mass / molar mass

molar mass of  Pb3(PO4)2 = 811.5427 (g/mol)

Number of moles of  Pb3(PO4)2 = 1.41 (g) / 811.5427 (g/mol) = 1.73 x 10^-3 mol

1 mole of Pb3(PO4)2 requires 6 moles of NaNO3

1.73 x 10^-3 mol of Pb3(PO4)2 requires 6 x 1.73 x 10^-3 mol

number of moles NaNO3 = 6 x 1.73 x 10^-3 mol = 10.38 x 10^-3 mol

Volume of Solution = 50 ml = 0.05 L

Molarity of NaNO3 = moles of Solute / Volume of solution in Litres = 10.38 x 10^-3 mol / 0.05 L

= 0.208 (mol/L)

b)

Given Chemical Equation

3KOH + H3Z---> K3Z + 3H2O.H3Z

volume of H3Z = 30 ml = 0.03 L

Molarity of H3Z = 0.0250 M

Number of Moles of H3Z = volume x molarity = (0.03 x 0.0250 ) mol = 7.5 x 10^-4 mol

1 mol of H3Z requires 3 moles of base KOH to completely neutralize

7.5 x 10^-4 mol of H3Z requires 3 x 7.5 x 10^-4 mol of KOH to completely neutralize

number of moles of KOH =  3 x 7.5 x 10^-4 mol = 2.25 x 10^-3 mol

molarity of KOH (base) = 0.105 M

Volume of base (KOH) in L = number of moles / molarity = 2.25 x 10^-3 mol / 0.105 M

= 0.0214 L

Volume of KOH = 21.43 ml requires to completely nwuteralise H3Z.