Class 11 Exam > Class 11 Questions > If Kspfor HgSO4is 6.4 × 10-5, then solub...
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If Ksp for HgSO4 is 6.4 × 10-5, then solubility of this substance in mole per m3 is
- a)8 × 10-3
- b)6.4 × 10-5
- c)8 × 10-6
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
If Kspfor HgSO4is 6.4 × 10-5, then solubility of this substance i...
Ksp = S2
⇒ 6.4 × 10–5 = S2
⇒ S = 8 × 10–3 mole/L
S = 8 × 10–3 × 103 mole/m3
⇒ S = 8 mole/m3
Most Upvoted Answer
If Kspfor HgSO4is 6.4 × 10-5, then solubility of this substance i...
S=root6.4*10^-5
=8*10^-3 mol/l
=8 mol/m^3
Hope this is helpful for you!!!!
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Community Answer
If Kspfor HgSO4is 6.4 × 10-5, then solubility of this substance i...
To determine the solubility of HgSO4 in moles per m3, we can use the equilibrium expression for the solubility product constant (Ksp). The solubility product constant is the product of the concentrations of the ions raised to the power of their coefficients in the balanced chemical equation.
The balanced chemical equation for the dissociation of HgSO4 is:
HgSO4(s) ⇌ Hg2+(aq) + SO42-(aq)
The solubility product constant expression for this equation is:
Ksp = [Hg2+][SO42-]
Given that the Ksp for HgSO4 is 6.4 × 10-5, we can use this information to calculate the solubility of HgSO4 in moles per m3.
Let's assume that the solubility of HgSO4 is represented by x. Since HgSO4 dissociates into one Hg2+ ion and one SO42- ion, the concentration of Hg2+ and SO42- ions in the equilibrium will also be x.
Therefore, we can write the expression for Ksp as:
Ksp = x * x = x2
Now, we can solve for x by taking the square root of Ksp:
x = √(Ksp) = √(6.4 × 10-5)
x ≈ 8 × 10-3
So, the solubility of HgSO4 in moles per m3 is approximately 8 × 10-3.
Therefore, the correct answer is option 'D' (None of these), as none of the given options match the calculated solubility.
The balanced chemical equation for the dissociation of HgSO4 is:
HgSO4(s) ⇌ Hg2+(aq) + SO42-(aq)
The solubility product constant expression for this equation is:
Ksp = [Hg2+][SO42-]
Given that the Ksp for HgSO4 is 6.4 × 10-5, we can use this information to calculate the solubility of HgSO4 in moles per m3.
Let's assume that the solubility of HgSO4 is represented by x. Since HgSO4 dissociates into one Hg2+ ion and one SO42- ion, the concentration of Hg2+ and SO42- ions in the equilibrium will also be x.
Therefore, we can write the expression for Ksp as:
Ksp = x * x = x2
Now, we can solve for x by taking the square root of Ksp:
x = √(Ksp) = √(6.4 × 10-5)
x ≈ 8 × 10-3
So, the solubility of HgSO4 in moles per m3 is approximately 8 × 10-3.
Therefore, the correct answer is option 'D' (None of these), as none of the given options match the calculated solubility.
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