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
Buffer capacity of a buffer solution is x, the volume of 1 M NaOH added to 100 mL of this solution if change the pH by 1...The base imidazole has a Kb of 1.0 × 10–7 at 25°C. In what volumes should 0.02 M-HCl and 0.02 M imidazole be mixed to ma...Graphite is :-...The hydronium ion concentration in an aqueous solution of H2SO4 is 2.0 × 10−4 M at 25o C. The hydroxide ion concentratio...Ratio of energy of photon of wavelength 3000 and 6000 is...