jika kedalam 100 mL larutan HCL 01 M ditambahkan 100 mL larutan NH3 0,1 M (kb NH3= 10-5) perubahan

Pertanyaan:

jika kedalam 100 mL larutan HCL 01 M ditambahkan 100 mL larutan NH3 0,1 M (kb NH3= 10-5) perubahan pH larutan tersebut adalah…

Larutan awal adalah 100 100 1 0 0 mL H C l cap H cap C l H C l 0 , 1 0 comma 1 0 , 1 M. Karena H C l cap H cap C l H C l adalah asam kuat, konsentrasi [ H + ] open bracket cap H raised to the positive power close bracket [ H + ] sama dengan konsentrasi asamnya. [ H + ] = [ H C l ] = 0 , 1 M = 10 -1 M open bracket cap H raised to the positive power close bracket equals open bracket cap H cap C l close bracket equals 0 comma 1 M equals 10 to the negative 1 power M [ H + ] = [ H C l ] = 0 , 1 M = 1 0 − 1 M p H a w a l = − log ( 10 -1 ) = 1 p cap H sub a w a l end-sub equals negative log open paren 10 to the negative 1 power close paren equals 1 p H a w a l = − l o g ( 1 0 − 1 ) = 1

Hitung jumlah mol masing-masing zat sebelum bereaksi:

Reaksi antara asam kuat ( H C l cap H cap C l H C l ) dan basa lemah ( N H 3 cap N cap H sub 3 N H 3 ) adalah sebagai berikut: N H 3 ( a q ) + H C l ( a q ) → N H 4 C l ( a q ) cap N cap H sub 3 open paren a q close paren plus cap H cap C l open paren a q close paren right arrow cap N cap H sub 4 cap C l open paren a q close paren N H 3 ( a q ) + H C l ( a q ) → N H 4 C l ( a q ) Karena mol kedua zat sama ( 10 10 1 0 mmol), keduanya habis bereaksi dan membentuk garam N H 4 C l cap N cap H sub 4 cap C l N H 4 C l sebanyak 10 10 1 0 mmol. Larutan ini mengalami hidrolisis garam asam.

Konsentrasi garam [ N H 4 C l ] open bracket cap N cap H sub 4 cap C l close bracket [ N H 4 C l ] dalam volume total ( 100 + 100 = 200 100 plus 100 equals 200 1 0 0 + 1 0 0 = 2 0 0 mL): [ M ] = 10 mmol 200 mL = 0 , 05 M = 5 × 10 -2 M open bracket cap M close bracket equals the fraction with numerator 10 mmol and denominator 200 mL end-fraction equals 0 comma 05 M equals 5 cross 10 to the negative 2 power M [ M ] = 1 0 m m o l 2 0 0 m L = 0 , 0 5 M = 5 × 1 0 − 2 M Gunakan rumus hidrolisis asam: [ H + ] = K w K b × [ M ] open bracket cap H raised to the positive power close bracket equals the square root of the fraction with numerator cap K sub w and denominator cap K sub b end-fraction cross open bracket cap M close bracket end-root [ H + ] = K w K b × [ M ] [ H + ] = 10 -14 10 -5 × 5 × 10 -2 = 5 × 10 -11 = 50 × 10 -12 = 7 , 07 × 10 -6 M open bracket cap H raised to the positive power close bracket equals the square root of the fraction with numerator 10 to the negative 14 power and denominator 10 to the negative 5 power end-fraction cross 5 cross 10 to the negative 2 power end-root equals the square root of 5 cross 10 to the negative 11 power end-root equals the square root of 50 cross 10 to the negative 12 power end-root equals 7 comma 07 cross 10 to the negative 6 power M [ H + ] = 1 0 − 1 4 1 0 − 5 × 5 × 1 0 − 2 = 5 × 1 0 − 1 1 √ = 5 0 × 1 0 − 1 2 √ = 7 , 0 7 × 1 0 − 6 M p H a k h i r = − log ( 7 , 07 × 10 -6 ) ≈ 5 , 15 p cap H sub a k h i r end-sub equals negative log open paren 7 comma 07 cross 10 to the negative 6 power close paren is approximately equal to 5 comma 15 p H a k h i r = − l o g ( 7 , 0 7 × 1 0 − 6 ) ≈ 5 , 1 5

Δ p H = | p H a k h i r − p H a w a l | delta p cap H equals the absolute value of p cap H sub a k h i r end-sub minus p cap H sub a w a l end-sub end-absolute-value Δ p H = | p H a k h i r − p H a w a l | Δ p H = 5 , 15 − 1 = 4 , 15 delta p cap H equals 5 comma 15 minus 1 equals 4 comma 15 Δ p H = 5 , 1 5 − 1 = 4 , 1 5

Perubahan pH larutan tersebut adalah 4,15 .