Valentine V
24 Mei 2023 01:50
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Valentine V
24 Mei 2023 01:50
Pertanyaan
Ke dalam 100ml larutan NaOH 0,1m ditambahkan 100ml larutan CH3COOH 0,1m. Jika Ka CH3COOH = 2×10^-5, ph campuran yg terjadi adalah
Ke dalam 100ml larutan NaOH 0,1m ditambahkan 100ml larutan CH3COOH 0,1m. Jika Ka CH3COOH = 2×10^-5, ph campuran yg terjadi adalah
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Gilangg G
24 Mei 2023 03:32
<p>Diketahui :</p><p>V NaOH = 100mL = 0,1 L</p><p>V CH3COOH = 100mL = 0,1 L</p><p>M NaOH = 0,1 M</p><p>M CH3COOH = 0,1 M</p><p>Ka CH3COOH = 2x10^-5</p><p> </p><p>Ditanya :</p><p>pH Campuran = ?</p><p> </p><p>Jawab</p><p>n NaOH = M/V</p><p>n NaOH = 0,1/0,1</p><p>n NaOH = 1 mol</p><p> </p><p>n CH3COOH = 0,1/0,1</p><p>n CH3COOH = 1 mol</p><p> </p><p> CH3COOH + NaOH --> CH3COONa + H2O</p><p>Mula" : 1 mol + 1 mol</p><p>Reaksi : -1 mol + -1 mol --> +1 mol</p><p>Sisa : - + - --> 1 mol</p><p> </p><p>Karena reaktan habis bereaksi dan hanya tersisa garam, maka digunakan konsep Hidrolisis Garam</p><p> </p><p>V Campuran (CH3COONa) = 100 mL + 100mL = 200mL = 0,2 L</p><p>M CH3COONa = n x V</p><p>M CH3COONa = 1 x 0,2 L</p><p>M CH3COONa = 0,2 M</p><p> </p><p>[OH-] = √(Kw/Ka) x [CH3COONa]</p><p>[OH-] = √(10^-14 / 2x10^-5) x 0,2</p><p>[OH-] = √5x10^-10 x 2x10^-1</p><p>[OH-] = √10^-10</p><p>[OH-] = 10^-5</p><p> </p><p>pOH = - log [OH-]</p><p>pOH = - log 10^-5</p><p>pOH = 5</p><p> </p><p>pH = 14 - pOH</p><p>pH = 14 - 5</p><p>pH = 9</p>
Diketahui :
V NaOH = 100mL = 0,1 L
V CH3COOH = 100mL = 0,1 L
M NaOH = 0,1 M
M CH3COOH = 0,1 M
Ka CH3COOH = 2x10^-5
Ditanya :
pH Campuran = ?
Jawab
n NaOH = M/V
n NaOH = 0,1/0,1
n NaOH = 1 mol
n CH3COOH = 0,1/0,1
n CH3COOH = 1 mol
CH3COOH + NaOH --> CH3COONa + H2O
Mula" : 1 mol + 1 mol
Reaksi : -1 mol + -1 mol --> +1 mol
Sisa : - + - --> 1 mol
Karena reaktan habis bereaksi dan hanya tersisa garam, maka digunakan konsep Hidrolisis Garam
V Campuran (CH3COONa) = 100 mL + 100mL = 200mL = 0,2 L
M CH3COONa = n x V
M CH3COONa = 1 x 0,2 L
M CH3COONa = 0,2 M
[OH-] = √(Kw/Ka) x [CH3COONa]
[OH-] = √(10^-14 / 2x10^-5) x 0,2
[OH-] = √5x10^-10 x 2x10^-1
[OH-] = √10^-10
[OH-] = 10^-5
pOH = - log [OH-]
pOH = - log 10^-5
pOH = 5
pH = 14 - pOH
pH = 14 - 5
pH = 9
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