The solubility product of PbI 2 is 7.47 × 10-9 at 15° C and 1.39 × 108 at 25° C. The molar heat of solution of PbI 2 is (use log 1.86=0.2695 )

Question

The solubility product of PbI 2 is 7.47 × 10-9 at 15° C and 1.39 × 108 at 25° C. The molar heat of solution of PbI 2 is (use log 1.86=0.2695 ) (A) 44.29 kJ/mol (B) 46.25 kJ/mol (C) 29.37 kJ/mol (D) 21.15 kJ/mol

Tardigrade Admin · Accepted Answer

log (( K sp )2/( K sp )2)=(Δ H /2.303 R )(( T 2- T 1/ T 1 T 2)) log (1.39 × 10-8/7.47 × 10-9)=(Δ H /2.303 × 8.314)((298-288/288 × 298)) Δ H =44.29 kJ / mole

Q. The solubility product of $P b I_{2}$ is $7.47 \times 1 0^{- 9}$ at $1 5^{\circ} C$ and $1.39 \times 1 0^{8}$ at $2 5^{\circ} C$ . The molar heat of solution of $P b I_{2}$ is (use log $1.86 = 0.2695$ )

44.29 kJ/mol

46.25 kJ/mol

29.37 kJ/mol

21.15 kJ/mol

$lo g \frac{( K _{s p} ) _{2}}{( K _{s p} ) _{2}} = \frac{Δ H}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$
$lo g \frac{1.39 \times 1 0 ^{- 8}}{7.47 \times 1 0 ^{- 9}} = \frac{Δ H}{2.303 \times 8.314} (\frac{298 - 288}{288 \times 298})$
$Δ H = 44.29 k J$ / mole

Q. The solubility product of PbI2​ is 7.47×10−9 at 15∘C and 1.39×108 at 25∘C. The molar heat of solution of PbI2​ is (use log 1.86=0.2695 )