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The Ka value given for NH4+ in my book is 5.70x10^-10.Calculate the pH of the solution that results when 42.0 mL of 0.1310 M NH3 is (This problem requires values in your textbook

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a) According to dilution formula Miv, = MV2 +0.1310x42.0 =4*80.0 0.1310 x 42.0 80.0 0688M 2 NH₃ + HO -* 0.0688 HO + oH - 0 0M2 V2 2) (NH47- My •e: 310 x 20 $9.687 3 4 (vN2) (42+21) 0.26221.0 50.0873 M (Hd = salt) (Vet V2) 7 42+ 21) pOH = pkb + log (

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