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Calculate either H,0or [OH-] for each of the solutions at 25 °C. Solution A: [OH-] = 1.81 x 10-7M Solution A: 1,0) = Solution
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According to the condition given in question Solution-A :- The ion Pzation constant of water at 25°C is given as fonows Kw =Solution - B Given [H₂ot ] = 9,85x10 M Kw = [H30+3[04] 1014 - 9.85 x169 com ] [OH-] = 10-14 9,85x10-9 [OH-]-[1,015 x 10 o/mSolution-c Given [H3073 € 0.000631 M Kw = [H3073[04] Kw = 10-14 1014 = 0.000631 [OH-] - [ok ] = 10-14 0.000631 Ti 58 x 10- [OpH = -log[H30+ ] Solution-A [H307 3 = 9.85x1099 PHE -log(9.85X169) pH = 8 solution - B [H30+3 = 0.000631 m pH = -log(0.000631CHIJ COH-] = 1014 H+ = 10-14 1,81x107 C#+3 = 5,52 x 108 pH -log(5.59x168) 1 pH = 7: 26. As for the solutions A and C, - the p

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