7.68. The solubility product constant of Ag2CrO4 and AgBr are 1.1×10−12and 5.0×10-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
7.68. The solubility product constant of Ag2CrO4 and AgBr are 1.1×10−12and 5.0×10-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
-
1 Answer
-
Silver chromate: Ag2CrO4? ? 2Ag+ + CrO42−?
[Ag+] = 2s1? , CrO42−? = s1?
Ksp? = (2s1? )2 (s1? ) = 4s3 = 1.1×10−12
s1? = 6.5 × 10−5 . (1)
Silver bromide: AgBr? Ag+ + Br−
[Ag+] = [Br−] = s2?
Ksp? = (s2? ) × (s2? ) = (s2)2? = 5.0 × 10−13
s2? =7.07×10−7. (2)
Divide equation (1) by equation (2) to obtain the ratio of the molarities of saturated solutions:
? s1/s2? = (6.50×10−5)/ (7.07×10−7)? = 91.9
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 0
Eqb. moles n(1 – a) na
total moles =
Eqb. pressure
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
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 687k Reviews
- 1800k Answers