Kirakan ["H" ^ +], ["OH" ^ -] dan "pH" penyelesaian 0.75 M "HNO" _2. (K_a = 4.5xx10 ^ -4)?

Kirakan ["H" ^ +], ["OH" ^ -] dan "pH" penyelesaian 0.75 M "HNO" _2. (K_a = 4.5xx10 ^ -4)?
Anonim

Jawapan:

# "H" ^ + = 0.0184mol # # dm ^ -3 #

# "OH" ^ - = 5.43 * 10 ^ -13mol # # dm ^ -3 #

# "pH" = 1.74 #

Penjelasan:

# K_a # diberikan oleh:

#K_a = ("H" ^ + "A" ^ -) / ("HA") #

Walau bagaimanapun, untuk asid lemah ini adalah:

#K_a = ("H" ^ + ^ 2) / ("HA") #

# "H" ^ + = sqrt (K_a "HA") = sqrt (0.75 (4.5xx10 ^ -4)) = 0.0184mol # # dm ^ -3 #

# "OH" ^ - = (1 * 10 ^ -4) /0.0184=5.43*10^-13mol # # dm ^ -3 #

# "pH" = - log ("H" ^ +) = - log (0.0184) = 1.74 #