Advertisements
Advertisements
Question
Ksp of AgCl is 1.8 × 10−10. Calculate molar solubility in 1 M AgNO3.
Advertisements
Solution
\[\ce{AgCl_{(s)} ⇌ Ag^+_{( aq)} + Cl^-_{( aq)}}\]
x = solubility of AgCI in 1 M AgNO3
\[\ce{AgNO3_{(aq)} ⇌ \underset{1 M}{Ag^+_{( aq)}} + \underset{1 M}{NO^-_{3(aq)}}}\]
[Ag+] = x + 1 ≃ 1 M ..........(∵ x << 1)
[Cl–] = x
Ksp = [Ag+] [Cl–]
1.8 × 10−10 = (1) (x)
x = 1.8 × 10−10 M
APPEARS IN
RELATED QUESTIONS
Answer the following :
Explain the relation between ionic product and solubility product to predict whether a precipitate will form when two solutions are mixed?
If ‘IP’ is the ionic product and ‘Ksp’ is the solubility product, precipitation of the compound will occur under the condition when:
The solubility product of BaCl2 is 4.0 × 10-8. What will be its molar solubility in mol dm-3?
Write solubility product of following sparingly soluble salt.
CaF2
The solubility of AgBr in water is 1.20 × 10–5 mol dm–3. Calculate the solubility product of AgBr.
Define Solubility product.
The solubility of AgCl(s) with solubility product 1.6 × 10−10 in 0.1 M NaCl solution would be ____________.
The solubility product of AgBr is 5.2 × 10−13 Calculate its solubility in mol dm−3 and g dm−3 (Molar mass of AgBr = 187.8g mol−1)
Calculate the current strength (I) and number of moles of electrons required to produce 2.369 × 10−3 kg of Cu from CuS04 in one hour. (Given: Molar mass of Cu = 63 .5 gm/mol).
The solubility of silver sulphate is 1.85 x 10-2 mol dm-3. Calculate solubility product of silver sulphate.
