Solution 0 200 grams of ammonium chloride is added to a 100 mL volumetric flask AFter dissolving 40 00 mL of a 0 09346 F LiOH solution is added What is the
Solution grams of ammonium chloride is added to a mL volumetric flask AFter dissolving mL of a F LiOH solution
Solution grams of ammonium chloride is added to a mL volumetric flask AFter dissolving mL
is added to a mL volumetric flask AFter dissolving mL of a F LiOH solution is added What is the
Solution grams of ammonium chloride is added to a mL volumetric flask
AFter dissolving mL of a F LiOH solution is added What is the
Solution grams of ammonium chloride is added to a mL
Solution grams of ammonium chloride
(Solution) 0.200 grams of ammonium chloride is added to a 100 mL volumetric flask. AFter dissolving, 40.00 mL of a 0.09346 F LiOH solution is added. What is the...

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0.200 grams of ammonium chloride is added to a 100 mL volumetric flask. AFter dissolving, 40.00 mL of a 0.09346 F LiOH solution is added. What is the pH of the final solution?Known values:Molar mass of NH4Cl: 53.491 g/molpKa = 9.25Ka = 1.73 x 10^-5Kb = 1.73 x 10^-50.00374 mol NH4Cl0.00374 mol LiOHAnswer: pH = 10.838I have calculated / obtained the known values provided above. I have also provided the answer for this problem. However, I am not sure how to solve the problem. All I know is that the -OH ions will react with NH4Cl to form ammonia. However, I am not sure how to get the pH from this.