A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Qip) becomes greater than its solubility product. If the aolubiltgy of BaSO4 in water is 8×10^-4 mol dm ^-3. Calculate its solubility in 0.01 mol dm ^-3 of H2SO4
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REFER THE ATTACHMENT GIVEN ABOVE
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Explanation:
Ksp for BaSO4 in water = [Ba2+] [ SO42- ] = (S) (S) = S2 But S = 8 × 10–4 mol dm–3 ∴ Ksp = (8×10–4)2 = 64 × 10–8 ... (1) The expression for Ksp in the presence of sulphuric acid will be as follows : Ksp = (S) (S + 0.01) ... (2) Since value of Ksp will not change in the presence of sulphuric acid, therefore from (1) and (2) Read more on Sarthaks.com - https://www.sarthaks.com/123968/sparingly-soluble-salt-gets-precipitated-only-when-product-concentration-ions-solution
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