Ionic EquilibriumHard
Question
The ionic product of water is 1.0 × 10−14 at 25o C. Assuming the density of water independent from change in temperature, the ionic product of water at 50o C will be
Options
A.2.0 × 10−14
B.5.0 × 10−15
C.5.9 × 10−14
D.1.0 × 10−14
Solution
$\ln\frac{K_{2}}{K_{1}} = \frac{\Delta H}{R}\left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$
$\text{Or, ln}\frac{K_{2}}{10^{- 14}} = \frac{13.7 \times 10^{3}}{2}\left( \frac{1}{298} - \frac{1}{323} \right) \Rightarrow K_{2} = 5.9 \times 10^{- 14}$
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
If K1 and K2 are the first and second ionization constants of H2CO3 and K1 >> K2, then the incorrect relation(s) i...10 mL of 10-6 M HCI solution is mixed with 90 mL H2O. pH will change approximately:...A cyclic process ABCD is shown in PV diagram for an ideal gas. Which of the following diagram represents the same proces...Which of the following weakest base ?...What will be the resultant pH when 200 ml of an aqueous solution of HCl (pH = 2.0) is mixed with 300 ml of an aqueous so...