BAB IV

HASIL PENGAMATAN

4.1 Data sample awal

ml sampel = 10 ml

ml KMnO4 (1) (a) = 5.4 ml

ml KMnO4 (2) = 12.5 ml

faktor ketelitian (f ) = 10

ml KMnO4 (2)

= 10

12.5

= 0.8

mg/L KMnO4 = 1000

mlsampelx {(10,0+a ) f−10 }x 0,01 x31,6

= 1000

10x {(10,0+5.4 ) 0.8−10 } x0,01 x31,6

= 73.312 mg/L

mg/lt KMnO4 = 73.312 / 3 = 24.33 mg/L

4.2 Pengenceran

Pengenceran

Volume sampel (ml) = volume botol sampel

P1

Volume pengencer (ml) = volume botol sampel – volume sampel

P1 = 24.33 mg/l

Volume labu Volume sample Volume pengencerBlangko0 332 0 332

Sampel0,1 250 10.4 239.6

Sampel0,2 325 13.5 311.5

Blangko7 315 0 315

Sampel7,1 305 12.7 292.3

Sampel7,2 321 13.4 307.6

4.1.3 Penetapan Oksigen Terlarut Metoda Winkler

Konsentrasi Thiosulfat (N) = 0.0125 N

Parameter ml Tiosulfat yang dibutuhkan untuk Titrasi

Botol Erlenmeyer Total Rata-rata

BOD0 Blanko 0 7.9 5.4 13.3 13.3

Sample 1 10.8 6.8 17.614.35

Sample 2 8.1 3 11.1

BOD 7 Blanko 7 10.2 2.9 13.1 13.1

Sample 1 12.1 4.9 17

Sample 2 7.6 6.2 13.8

4.1.4 Penentuan BOD

mg/L O2 =1000x mlTiosulfat x N x 8mlVolume Botol−2ml

Parameter mg/L O2

DO 0 (A) 5.26665335

DO 7 (B) 4.96828993

Blanko 0 (C) 4.0303

Blanko 7 (D) 4.1853

BOD = P ( A - B ) – ( C – D )

= 3 (5.26665335 – 4.96828993) – (4.0303 – 4.1853)

= 1.05009 mg/L

LAMPIRAN

1. Lampiran Perhitungan

- Pengenceran

1. Blangko0

Volume sampel (ml) = 0 ml

Volume pengencer (ml) = 332 ml

2. Sampel0,1

Volume sampel (ml) = 250

24.04 = 10.4 ml

Volume pengencer (ml) = volume botol sampel – 10.4

= 239.6

3. Sampel0,2

Volume sampel (ml) = 32 5

24.04 = 13.5 mL

Volume pengencer (ml) = volume botol sampel – 13.5 mL

= 311.5

4. Blangko7

Volume sampel (ml) = 0 ml

Volume pengencer (ml) = 315 ml

5. Sampel7,1

Volume sampel (ml) = 305

24.04 = 12.7 ml

Volume pengencer (ml) = volume botol sampel – 12.7 mL

= 292.3 ml

6. Sampel7,2

Volume sampel (ml) = 321

24.04 = 13.4 ml

Volume pengencer (ml) = volume botol sampel – 13.04

= 307.6 ml

Perhitungan BOD

BOD = P (A-B) – (C-D)

A 01 = mg/l O2 sampel nol hari

=1000x 17.6 x1/80 x 8

250−2ml = 7.09674 mg/L O2

A 02 = mg/l O2 sampel nol hari

=1000x 11,1 x1/80 x 8

325−2ml = 3.436533 mg/L O2

A rata-rata =7.09674+3.346533

2= 5.26665335 mg/L O2

B = mg/l O2 sampel tujuh hari

B01 =1000x 17 x1/80 x 8

305−2ml = 5.610561 mg/L O2

B 02 =1000x 13.8 x1/80 x8

321−2ml = 4.326109 mg/L O2

Rata-rata B= 4.9682 mg/L O2

C = mg/l O2 blangko nol hari

1000x 13.3 x 1/80 x8332−2ml

= 4.0303 ml

D = mg/l O2 blangko tujuh hari

1000x 13,1x 1/80 x8315−2ml

= 4.1853 ml

`

BOD = P ( A - B ) – ( C – D )

= 3 (5.2666– 4.9682) – ( 4.0304 – 4.1853 )

= 1.05009 mg/L

Selisih pengurangan DO7 dengan DO0

Selisih pengurangan

% Selisih pengurangan = nilai DO0 sampel−DO7 sampel

nilai DO0×100 %

% Selisih pengurangan = 5.2666

mgL

−4.9682mgL

5.2666mg /L×100 %

% Selisih pengurangan = 5.66 %