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Answer to Question #16280 in Inorganic Chemistry for nontsi

Question #16280
calculate the pH of the solution that contains 4.25g NH3 and 3.65g HCl per 500cm3 of the solution. Kb of NH3=1.8*10^-5
Expert's answer
n of NH3 = 4.25/17=0.25 mol
n of HCL=3.65/36.5=0.1 mol

0.25 & 1-0.25 0.25
NH3& +& HC & l= NH4Cl

we have 500 ml with 0.75 mol HCl and 0.25 mol of NH4Cl, concentration of HCl = [1.5 ], and NH4Cl= [0.5]

pH=-logKb + [NH4Cl]/[HCl]& = 4.74 +0.33 =5.07

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