100 ml larutan CH3COOH 0,2 M dicampurkan dengan 100 ml larutan NaOH 0,1 M (Ka CH3COOH=10 pangkat -5) . Tentukan pH larutan sebelum dan sesudah dicampurkan!
fiqahdwita
CH₃COOH + NaOH ⇒ CH₃COONa + H₂O M 20 10 - - T 10 10 10 10 S 10 - 10 10 Terdapat sisa basa lemah mol basa sisa = 10 mmol [H⁺] = Ka × a/g 10⁻⁵ × 10 / 10 = 10 ⁻⁵ ph = 5- log 1 ph = 5
jika salah mohon dimaafkan .
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Prasetya
Dik: [CH₃COOH] = 0,2 M | 100ml (Ka = 10⁻⁵) [NaOH] = 0,1 M | 100ml
pH sebelum Reaksi berarti kita hitung pH masing masing larutan: pH CH₃COOH ==> [H⁺] = [H⁺] = [H⁺] = 10⁻³ x √2 pH = - log [H⁺] pH = - log 10⁻³ x √2 pH = 5 -log √2
pH NaOH ==> [OH⁻] = b x [NaOH] [OH⁻] = 1 x 0,1 M = 0,1 M pH = 14 -(-log [OH⁻]) H = 14-(-log 0,1) pH = 13
CH₃COOH + NaOH ==> CH₃COONa + H₂O Mula 20 mmol 10 mmol Reaksi -10 mmol -10 mmol +10 mmol +10 mmol Sisa 10 mmol - 10 mmol 10 mmol Terdapat sisa Asam lemah, yang berarti larutan ini bersifat Penyangga, maka untuk mencari pH : [H⁺] = Ka x [H⁺] = 10⁻⁵ x [H⁺] = 10⁻⁵
M 20 10 - -
T 10 10 10 10
S 10 - 10 10
Terdapat sisa basa lemah
mol basa sisa = 10 mmol
[H⁺] = Ka × a/g
10⁻⁵ × 10 / 10 = 10 ⁻⁵
ph = 5- log 1
ph = 5
jika salah mohon dimaafkan .
[CH₃COOH] = 0,2 M | 100ml (Ka = 10⁻⁵)
[NaOH] = 0,1 M | 100ml
pH sebelum Reaksi berarti kita hitung pH masing masing larutan:
pH CH₃COOH ==> [H⁺] =
[H⁺] =
[H⁺] = 10⁻³ x √2
pH = - log [H⁺]
pH = - log 10⁻³ x √2
pH = 5 -log √2
pH NaOH ==> [OH⁻] = b x [NaOH]
[OH⁻] = 1 x 0,1 M = 0,1 M
pH = 14 -(-log [OH⁻])
H = 14-(-log 0,1)
pH = 13
CH₃COOH + NaOH ==> CH₃COONa + H₂O
Mula 20 mmol 10 mmol
Reaksi -10 mmol -10 mmol +10 mmol +10 mmol
Sisa 10 mmol - 10 mmol 10 mmol
Terdapat sisa Asam lemah, yang berarti larutan ini bersifat Penyangga, maka untuk mencari pH :
[H⁺] = Ka x
[H⁺] = 10⁻⁵ x
[H⁺] = 10⁻⁵
pH = -log [H⁺]
pH = - log 10⁻⁵
pH = 5
Semoga Membantu!!
Terbaiknya!!!