Q2. (a) Define pH and Calculate hydrogen ion [H+] and hydroxyl ion [OH-] concentration for a rainwater sample with pH 5.6.

(5 marks)

(b) (i) Define the BOD520C.

(ii) If the BOD520C of a waste is 220.0 mg/L and the ultimate BOD, Lo is 320.0 mg/L,

what is the rate constant, k? (5 marks)

Q2. (a) Berikan definisi pH dan kira kepekatan ion hydrogen [H+] dan ion hydroxyl ion [OH-] bagi suatu sampel air hujan yang mempunyai pH 5.6.

(b) (i) Berikan definisi BOD520C.

(ii) sekiranya nilai BOD520C bagi suatu sisa adalah sebanyak 220.0 mg/L dan BOD

muktamad, Lo adalah sebanyak 320.0 mg/L, apakah nilai pemalar kadar tindakbalas, k?

ANSWER SCHEME:

Q2. (a) Define pH and Calculate hydrogen ion concentration [H+] for a rainwater sample with pH 5.6.

(5 marks)- Solution: pH is a measure of the concentration of hydrogen ions in water. - Acidity or alkaline of water. (2m)

pH = -log [H+] 5.6 = -log [H+]

Therefore, [H+] = antilog -5.6 = 10 -5.6 mol/L (1.5m)

pH + pOH = 14pOH = 14-5.6 = 8.4

pOH = -log [OH-]

8.4 = -log [OH-] Therefore, [OH-] = antilog -8.4

= 10 -8.4 mol/L (1.5m)

(b) (i) Define the BOD520C.

(2 marks)

pengukuran permintaan oksigen yang diperlukan oleh mikroorganisma bagi pengoksidaan

bahan organic yang boleh terurai (biodegradable organic matter) di dalam air sisa di dalam

keadaan aerobic selepas 5 hari pada suhu 20C.

((ii) If the BOD520C of a waste is 220.0 mg/L and the ultimate BOD, Lo is 320.0 mg/L, what

is the rate constant, k?

Solution:

a. use BOD rate equation

BODt= Lo (1– e-kt)

2m

220 = 320(1 - e -k(5) ) 0.6875 = 1 – e -k(5) -0.3125 = -e -k(5) b. Divide through by -1 and take natural logarithm of both sides ln(0.3125) = ln(e -k(5) ) -1.1632 = -k(5) k = 0.1843 d -1 3m