Question

NaCl(s) ⇌ Na+(aq) + Cl-(aq) ΔHo = 3.9 kJ/mol At 298 K, a saturated solution of NaCl has [Na+] = 7.0 M and [Cl-] = 5.4 M. If the temperature of the mixture is increased to 323 K, what will be the equilibrium concentration (M) of Na+? (Assume no ion pairing.) Enter your answer to 2 decimal places.

NaCl(s) Na (aq) + CI'(aq) AH- 3.9 kJ/mol At 298 K, a saturated solution of NaCI has [Na"] 7.0M and [CI] - 5.4 M. If the temperature of the mixture is increased to 360 K. what will be the equilibrium concentration (M) of Na ? (Assume no ion pairing.) Enter your answer to 2 decimal places.

Answer 1

Answer:

Given,

To determine the equilibrium concentration of Na+

Let us consider,

Reaction is given below

i.e.,

NaCl $\rightleftharpoons$ Na+ + Cl-

Equilibrium constant Kc = [Na+][Cl-]

= 7*5.4

Kc = 37.8

DeltaG = - RTlnK

= - 8.314 * 298 * ln K

consider,

- RT ln K = DeltaH - TDeltaS

ln K = - DeltaH/RT + DeltaS/R

ln K1 - ln K2 = - DeltaH/R *(1/T1 - 1/T2)

substitute values

ln 37.8 - ln K2 = - 3.9*10^3/8.314 *(1/298 - 1/334)

On solving we get

ln K2 = 3.46

K2 = 31.81

Now we have to consider,

[Na+][Cl-] = 31.81

It is given that [Cl-] = 5.4

[Na+] = 31.81 / 5.4

Concentration of Na+

i.e.,

[Na+] = 5.89 M