0.25 M solution of pyridinium chloride C₅H₆N⁺Cl⁻ was found to have a pH of 2.699. if the value of Kₑ of pyridine C₅H₅N is a * 10⁻¹⁰, then the value of a is (given, 10⁻².⁶⁹⁹ = 2 * 10⁻³)
0.25 M solution of pyridinium chloride C₅H₆N⁺Cl⁻ was found to have a pH of 2.699. if the value of Kₑ of pyridine C₅H₅N is a * 10⁻¹⁰, then the value of a is (given, 10⁻².⁶⁹⁹ = 2 * 10⁻³)
BCl → B? + Cl?
B? + H? O? BOH + H? , K? = K? /K?
0.25 – –
(0.25 – x) x
Given, pH = 2.7 = [H? ] = 2 * 10? ³
∴ x²/ (0.25) = (10? ¹? )/K?
= 4 * 10? * 4 * K? = 10? ¹?
= K? = (1/16) * 10? = 6.25 * 10? ¹?
Similar Questions for you
0.01 M NaOH,
M = 1 * 10-2

pOH = 2
pH = 2
Kp = Kc (RT)Dng
36 * 10–2 = Kc (0.0821 * 300)–1
Kc = 0.36 * 0.0821 * 300 = 8.86 » 9
A(g) ->B(g) + (g)
Initial moles n 0 &nbs
On increasing pressure, equilibrium moves in that direction where number of gaseous moles decreases.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
Learn more about...

Chemistry Chemical Equilibrium 2025
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
See what others like you are asking & answering

