A student mixes 51.0 mL of 2.98 M Pb(NO3)2(aq) with 20.0 mL of 0.00243 M Na2C2O4

Question

A student mixes 51.0 mL of 2.98 M Pb(NO3)2(aq) with 20.0 mL of 0.00243 M Na2C2O4(aq). How many moles of PbC2O4(s) precipitate from the resulting solution? What are the values of [Pb2 ], [C2O42–], [NO3–], and [Na ] after the solution has reached equilibrium at 25 °C?

mstudent mixes 51.0 mL of 2.98 M Pb(NO3)2(aq) with 20.0 mL of 0.00243 M Na2C204(aq). How many moles of PbC204(s) precipitate from the resulting solution? Number KsJPbC,04(s)|-8.5× 10-9 M2 mol sp What are the values of [Pb2 [C2041, (NO3, and [Na] after the solution has reached equilibrium at 25 °C? Number Number Number Number

Explanation / Answer

Moles of  Pb(NO3)2(aq) present = 0.051*2.98 = 0.15196 moles

Moles of Na2C2O4(aq) present = 0.02 * 0.00243 = 4.86*10^-5 moles

Ksp of PbC2O4 = 8.5*10^-9

since Na2C2O4 is the limiting reagent , moles of PbC2O4(s) formed = 4.86*10^-5 moles

So now the concentration of moles / total volume

Pb+2 = 0.15196- 4.86*10^-5/ 0.071 = 2.128 M

NO3- = 2*0.15196 / 0.071 = 4.280 M

Na+ = 2*4.86*10^-5/0.071 = 1.369*10^-3 M

.PbC2O4 ==> Pb^2+ + C2O4^2-

Ksp = 8.5*10^-9 = (Pb+2)(C2O42-)

= (2.128+x) * x = 8.5*10^-9

= 3.994 *10^-9 M

Therefore Pb+2 = 2.128 M (approx)

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