Ionic EquilibriumMedium
Question
The solubility of AgCl in water is 0.001435 g per litre at 15o C. The solubility product of AgCl is
(Ag = 108, Cl = 35.3)
Options
A.10−5
B.10−10
C.2 × 10−10
D.10−9
Solution
$S = \frac{0.001435}{143.5} = 10^{- 5}\text{ M}$
∴ Solubility product, Ksp = S2 = 10–10
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Titration curves for 0.1M solutions of three weak acids HA1, HA2 and HA3 with ionization constants K1, K2 and K3 respect...One mole of CH3COOH and one mole of C2H5OH reacts to produce mol of CH3COOC2H2. The equilibrium constant is:...When 0.1 mole solid NaOH is added in 1lt of 0.1 M NH3(aq) then which statement is wrong? (Kb = 2 × 10-5, log 2 = 0....Which of the following are true for an acid-base titration ?...A 0.1 M solution of [Cu(NH3)4]+ is stirred with an excess of potassium cyanide sufficient to convert all the ammonium co...