Map General Chemistry 4th Edition University Science Books presented by Sapling

Question

Map General Chemistry 4th Edition University Science Books presented by Sapling Learning A student mixes 49.0 mL of 3.48 M Pb(NO3)2(aq) with 20.0 mL of 0.00243 M Nal(aq) How many moles of Pbl2(s) precipitate from the resulting solution? Number KsJPb12(s) = 9.8 × 10-9 4.86x 10-5mol What are the values of [Pb2], [I], [NO3], and [Na] after the solution has reached equilibrium at 25 °C? Number Number Pb2+ |-| 2.47 I3.97x 109M Number Number [Na+ 110.0000141- NO,- 4.94 Previous Try Again Next Exit Explanation

Explanation / Answer

moles of PbI2 theoretically:

mmol of Pb = MV = 49*3.48 = 170.52

mmol of I = MV = 20*0.00243 = 0.0486 mmol

clearly, I- is limiting

2 mol of I= 1 mol of PbI2

0.0486 mmol --> 0.0486/2 = 0.0243 mmol of PbI = 0.0243*10^-3 = 2.43*10^-5 mol of PbI2 will be formed

your error --> raito of I- vs. PbI2

Q2.

then...

I- in solution -> Form equilibrium

Ksp = [Pb+2][I-]^2

(9.8*10^-9) = (2.47) *[I-]^2

[I-] = ((9.8*10^-9)/2.47)^0.5

[I-] = 0.000062988 M = 6.29*10^-5 M

for Na+

[Na+] = mmol of NA+ /Vtotal = (0.0486 )/(49+20) = 0.0007043M

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.