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The pH at which Mg(OH)_2 [K_sp=1times 10^-11] begins to precipitate from a solution containing 0.10text M Mg^2+ ions is

Numerical Answer Type:
Enter a numerical value Answer: 9 to 9 +4 marks

Solution & Explanation

### Related Formula K_sp = [Mg^2+][OH^-]^2 pOH = -log[OH^-] pH + pOH = 14 ### Core Logic Precipitation begins just when the ionic product equals the solubility product (Q_sp = K_sp). ### Step 1: Calculating required [OH-] [Mg^2+][OH^-]^2 = 10^-11 Given [Mg^2+] = 0.10 text M 0.10 times [OH^-]^2 = 10^-11 [OH^-]^2 = 10^-10 [OH^-] = 10^-5 text M ### Step 2: Finding pH pOH = -log(10^-5) = 5 pH = 14 - pOH pH = 14 - 5 = 9 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

Reference Study Guides

More Equilibrium Previous-Year Questions — Page 6

Q62 jee_main_2024_31_jan_morning Equilibrium Constant
For the given reaction, choose the correct expression of K_C from the following :- Fe_(aq)^3+ + SCN_(aq)^- rightleftharpoons (FeSCN)_(aq)^2+
  • A. K_C = frac[FeSCN^2+][Fe^3+][SCN^-]
  • B. K_C = frac[Fe^3+][SCN^-][FeSCN^2+]
  • C. K_C = frac[FeSCN^2+][Fe^3+]^2[SCN^-]^2
  • D. K_C = frac[FeSCN^2+]^2[Fe^3+][SCN^-]

Solution

### Related Formula K_C = frac[textProducts][textReactants] ### Core Logic K_C = fractextProducts ion conc.textReactants ion conc. K_C = frac[FeSCN^2+][Fe^3+][SCN^-] ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

More Equilibrium Questions — jee_main_2024_30_jan_morning

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