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1. Penentuan larutan HCl ± 0,1 N dengan Na2CO3 anhidrat sebagai baku

Na2CO3 = 0,5260 gram

V Na2CO3 = 100 mL

V1 = 21 mL

V2 = 21 mL

V3 = 21 mL

1. Erlenmeyer 1

Mol Na2CO3 = 0,526 gram

106gram /mol = 0,00496 mol = 4,96 mmol

M = 4,96mmol

100mL = 0,0496 M

N = n x M = 2 x 0,0496 M = 0,0992 N

Molek Na2CO3 = Molek HCl

V x N = V1 x N1

25 mL x 0,0992 N = 21 mL x N1

2,48mLN

21mL = N1

0,118 N = N1

2. Erlenmeyer 2


106gram /mol = 0,00496 mol = 4,96 mmol

M = 4,96mmol

100mL = 0,0496 M

N = n x M = 2 x 0,0496 M = 0,0992 N


V x N = V2 x N2

25 mL x 0,0992 N = 21 mL x N2

2,48mLN

21mL = N2

0,118 N = N2

3. Erlenmeyer 3


106gram /mol = 0,00496 mol = 4,96 mmol

M = 4,96mmol

100mL = 0,0496 M

N = n x M = 2 x 0,0496 M = 0,0992 N


V x N = V3 x N3

25 mL x 0,0992 N = 21 mL x N3

2,48mLN

21mL = N3

0,118 N = N3

Jadi, konsentrasi rata-rata HCl yaitu 0,0496 M

2. Menentukan kadar NH3 dalam pupuk ZA(NH4)2SO4 = 0,1 gram = 100 mg

0,1 N NaOHNaOH mula-mula = N x V

= 0,1 x 10 = 1 mmol

a. Titrasi 1molek NaOH = molek HCl

= V x N = 2,2 mL x 0,118 N = 0,2596 mmol

mmol NaOH yang bereaksi = 1 mmol - 0,2596 mmol = 0,7404 mmol

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

(NH4)2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + NH3(g) + H2O(l)

mmol NH3 = 2mol NH 3

2mol NaOH x 0,7404 mmol = 0,7404 mmol

gram = 0,7404 mmol x 17 gr/mol = 12,5868 mg

% NH3 = 12,5868mg

100x 100% = 12,5868 %

b. Titrasi 2molek NaOH = molek HCl

= V x N = 2,3 mL x 0,118 N = 0,2714 mmol





% NH3 = 12,3862mg

100x 100% = 12,3862 %

c. Titrasi 3 molek NaOH = molek HCl

= V x N = 3,7 mL x 0,118 N = 0,4366 mmol





% NH3 = 9,5778mg

100x 100% = 9,5778 %

% rata-rata NH3 = 12,5868 %+12,3862 %+9,5778 %

3

= 34,5508 %

3 = 11,5169 %

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