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What is [HS^(-)] & [S^(2-)] in solution ...

What is `[HS^(-)] & [S^(2-)]` in solution of `0.01M HCl` and `0.1 M H_(2)S` ?(Given that for `H_(2)S,K_(a_(1))=10^(-7)and K_(a_(2))=10^(-14))`

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To find the concentrations of \([HS^-]\) and \([S^{2-}]\) in a solution of \(0.01M \, HCl\) and \(0.1M \, H_2S\), we can follow these steps: ### Step 1: Determine the concentration of \(H^+\) ions from \(HCl\) Since \(HCl\) is a strong acid, it completely dissociates in solution: \[ HCl \rightarrow H^+ + Cl^- \] Given that the concentration of \(HCl\) is \(0.01M\), the concentration of \(H^+\) ions from \(HCl\) is: ...
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Calculate pH_(1),[HS^(-)],[S^(2-)],[Cl^(-)] in a solution which is 0.1M in HCl& 0.1M in H_(2)S .Given that K_(a_(1))(H_(2)S)=10^(-7),K_(a_(2))(H_(2)S)=10^(-14) .Also calculate degree of dissociation of H_(2)S & HS^(-) in solution.

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RESONANCE ENGLISH-IONIC EQUILIBRIUM-Exercise -1
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  13. pH of urine is 6.0 . How many grams equivalents of the acid are elimin...

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  15. Calculate the pH of 0.2M sodium butyrate, (K(a) for butyric acid is 2....

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  16. A 0.25 M solution of pyridinium chloride C(5)H(5)NH^(+)Cl^(-) was foun...

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