The solving of problem on page 333

The solving of problem on page 333

1. Data

HCl : V = 50 mL

M = 0,1 M

NaOH : M = 0,1 M

HCl dititrasi dengan NaOH

Question

pH larutan :

a. sebelum dititrasi ?

b. setelah ditambah 25 mL NaOH 0,1 M ?

c. setelah ditambah 50 mL NaOH 0,1 M ?

d. setelah ditambah 100 mL NaOH 0,1 M ?

Answer

a. pH HCl = -log 0,1 = ……….

b. Mol HCl = 50 mL x 0,1 M = 5 mmol

Mol NaOH = 25 mL x 0,1 M = 2,5 mmol

	HCl	+	NaOH	®	NaCl	+	H2O
Mula-mula	5 mmol		2,5 mmol
Bereaksi	2,5 mmol		2,5 mmol
Sisa	2,5 mmol		0

Molaritas HCl =2,5 mmol / 75 mL = 0,033 M

pH = -log [HCl] = -log 0,033 = …………….

Jadi pH larutan = …………………………….

c. Mol HCl = 50 mL x 0,1 M = 5 mmol

Mol NaOH = 50 mL x 0,1 M = 5 mmol

	HCl	+	NaOH	®	NaCl	+	H2O
Mula-mula	5 mmol		5 mmol
Bereaksi	5 mmol		5 mmol
Sisa	0		0

Jadi pH larutan = ………………………..

d. Mol HCl = 50 mL x 0,1 M = 5 mmol

Mol NaOH = 100 mL x 0,1 M = 10 mmol

	HCl	+	NaOH	®	NaCl	+	H2O
Mula-mula	5 mmol		10 mmol
Bereaksi	5 mmol		5 mmol
Sisa	0		5

Molaritas NaOH = 5 mmol/150 mL = 0,033 M

pOH = ……………………………….

pH = ………………………………

Jadi pH larutan = ………………….

2. Data

NaOH : pH = 11

H₂SO₄ : volume = 100 mL

pH = 3

NaOH dicampurkan dengan H₂SO₄

pH campuran = 10

Question

Volume NaOH = ?

Answer

pH _NaOH = 11 ® pOH _NaOH = 14 – 11 = 3

pOH _NaOH = 3 ® [OH-] = 10-3 M

NaOH ® Na+ + OH-

[NaOH] = 1/1 x 10-3 M = 10-3 M

misal volume NaOH = x mL

maka mol NaOH = x mL x 10-3 M = 10-3x mmol

pH H₂SO₄ = 3 ® [H+] = 10-3 M

H₂SO₄ ® 2H+ + SO₄2-

[H₂SO₄] = ½ x 10-3 M = 0,5 x 10-3 M

mol H₂SO₄ = 100 mL x 0,5 x 10-3M = 50 x 10-3 mmol

	2NaOH	+	H2SO4	®	Na2SO4	+	2H2O
Mula-mula	x x 10-3 mmol		50 x 10-3 mmol
Bereaksi	100 x 10-3mmol		50 x 10-3 mmol
Sisa	(x – 100) x 10-3mmol		0

pH _campuran = 10 ® pOH _campuran = 14 – 10 = 4

pOH _campuran = 14 – 10 = 4 ® [OH^-] = 10-4 M

NaOH ® Na+ + OH^-

[NaOH] = 1/1 x 10-4 M = 10-4 M

volume campuran = mol NaOH / molaritas campuran

(100 + x ) mL = (x – 100) x 10-3mmol /10-4

(100 + x ) = (x – 100) x 10

(100 + x ) = 10 x – 1000

-10 x + x = -100 – 1000

-9 x = -1100

x = -1100/-9

x = 122,22

Jadi volume NaOH = 122,22 mL

3. Data

NH₄OH V = 500 mL

M = 0,1 M

Kb = 1x10-5

NH₄OH dicampurkan dengan (NH₄)₂SO₄

pH campuran = 9

Question

Massa (NH₄)₂SO₄ = ?

Answer

mol NH₄OH = 500 mL x 0,1 M = 50 mmol

pH campuran = 9 ® pOH = 14 - 9 = 5

[OH-] dalam campuran = 10-5

mol (NH₄)₂SO₄ = 50 mmol = 0,05 mol

g (NH₄)₂SO₄ = 0,05 mol x 132 g/mol = 6.6 g

jadi massa (NH₄)₂SO₄ yang harus ditambahkan = 6,6 g

4. Find yourself