Hardness Titration, the BOD Blank, and BOD of Raw Sewage: 18-Env-A4 Water and Wastewater Engineering, May 2013, Q2

Question 2: Hardness Titration, the BOD Blank, and BOD of Raw Sewage (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Hardness by EDTA Titration

The standard method (EDTA titrimetric method, e.g. Standard Methods 2340C) measures hardness — the sum of polyvalent cations, dominated by Ca²⁺ and Mg²⁺ — by titrating a buffered (pH ≈ 10) sample with a standardized solution of EDTA (ethylenediaminetetraacetic acid, a hexadentate chelating agent) in the presence of an Eriochrome Black T indicator. EDTA forms a strong, colourless 1:1 complex with each Ca²⁺ or Mg²⁺ ion; while free Ca²⁺/Mg²⁺ remains, the indicator is bound to the metal and the solution is wine-red. As titrant is added, EDTA preferentially strips the metal away from the weaker indicator complex; at the end point every free metal ion has been sequestered, the indicator reverts to its unbound (blue) colour, and the volume of titrant delivered is directly proportional to the total Ca+Mg equivalents originally present.

Given.

Given data
Quantity	Symbol	Value
Sample volume titrated	$V_s$	50 mL
Titrant volume to end point	$V_t$	5 mL
EDTA titrant normality	$N$	N/50 = 0.02 eq/L

Find. The total hardness of the water sample, in mg/L as CaCO₃.

Approach. Convert the titrant delivered to equivalents, convert equivalents to an equivalent mass of CaCO₃ using its equivalent weight (50 g/eq), then express that mass per litre of the original sample.

Equivalents of EDTA delivered at the end point. Because EDTA complexes Ca²⁺/Mg²⁺ 1:1 by charge, equivalents of titrant equal equivalents of hardness-causing cations in the aliquot: $$n_{eq}=V_t\times N=(5\times10^{-3}\ \text{L})\times(0.02\ \text{eq/L})=1.0\times10^{-4}\ \text{eq}.$$
Convert to an equivalent mass of CaCO₃. CaCO₃ has molar mass 100 g/mol and is divalent, so its equivalent weight is $100/2=50$ g/eq: $$m_{\text{CaCO}_3}=n_{eq}\times 50\ \text{g/eq}\times1000\ \text{mg/g}=1.0\times10^{-4}\times50\times1000=5.0\ \text{mg (in the 50 mL aliquot)}.$$
Express per litre of sample. $$\text{Hardness}=\frac{m_{\text{CaCO}_3}}{V_s}=\frac{5.0\ \text{mg}}{0.050\ \text{L}}=\boxed{100\ \text{mg/L as CaCO}_3}.$$

Check: this equals the familiar field shortcut for N/50 EDTA (chosen historically so 1 mL titrant ≈ 1 mg CaCO₃ per 100 mL sample) scaled to a 50 mL aliquot — both routes agree at 100 mg/L, which by the standard classification (0–75 soft, 75–150 moderately hard, 150–300 hard, >300 mg/L very hard as CaCO₃) places this water at the soft/moderately-hard boundary.

Quantity	Value
Equivalents of EDTA delivered	1.0×10^-4 eq
Equivalent mass of CaCO₃ (50 mL aliquot)	5.0 mg
Total hardness	100 mg/L as CaCO₃ (soft/moderately-hard boundary)

(b) Importance of the "Blank" in the Standard BOD Test

The blank (also called the seed/dilution-water control) is a bottle prepared identically to the sample bottles — same dilution water, same seed inoculum if used — but with no wastewater sample added. It is incubated alongside the test bottles for the same 5 days at 20°C and its dissolved-oxygen drop is measured exactly the same way. Its purpose is to isolate the oxygen demand contributed by the dilution water and seed organisms themselves (respiration of the seed microbes, and any residual organic/reducing matter in the dilution water) from the oxygen demand the test is actually trying to measure — that of the wastewater sample. Without the blank correction, every measured BOD would be biased high by whatever background depletion the dilution water and seed contribute on their own; Standard Methods requires the blank depletion not exceed about 0.2 mg/L over 5 days for the dilution water to be considered acceptable, and the blank's DO drop ($B_1-B_2$) is subtracted from the sample bottle's drop before the result is divided by the dilution fraction, exactly as done in part (c) below.

(c) BOD5 of the Undiluted Raw Sewage

Given.

Given data
Quantity	Symbol	Value
Sample fraction (dilution)	$P$	2% = 0.02
Initial DO, diluted sample & blank (same dilution water)	$D_1,\,B_1$	8.0 mg/L
Final DO (day 5), diluted sample	$D_2$	3.0 mg/L
Final DO (day 5), Blank	$B_2$	7.8 mg/L

Find. The 5-day BOD of the undiluted raw sewage, $\text{BOD}_5$.

Approach. Subtract the blank's own DO depletion from the diluted sample's DO depletion to isolate the demand due to the sewage alone, then scale up from the small percentage actually present in the bottle to the undiluted sewage by dividing by the sample fraction $P$.

DO depletion in the diluted sample bottle over 5 days. $$\Delta D=D_1-D_2=8.0-3.0=5.0\ \text{mg/L}.$$
DO depletion in the Blank over 5 days (background demand of the dilution water/seed). $$\Delta B=B_1-B_2=8.0-7.8=0.2\ \text{mg/L}.$$
Net oxygen consumed that is attributable to the sewage itself, per litre of the diluted (2%) bottle. $$\Delta D-\Delta B=5.0-0.2=4.8\ \text{mg/L}.$$
Scale from the 2% dilution up to the undiluted sewage. Since only a fraction $P$ of every litre in the bottle was actual sewage, the sewage's own concentration of demand is $1/P$ times larger: $$\text{BOD}_5=\frac{\Delta D-\Delta B}{P}=\frac{4.8}{0.02}=\boxed{240\ \text{mg/L}}.$$

Check: the blank's initial DO is assumed equal to the diluted sample's initial DO (8.0 mg/L) because both bottles are aerated from the same dilution-water source immediately before incubation — the question gives only one "start of test" value, consistent with this standard setup. A raw (undiluted) BOD₅ of 240 mg/L is squarely in the typical range for medium-strength untreated domestic sewage (110–350 mg/L, Metcalf & Eddy Table 3-15), which supports the dilution assumption.

Quantity	Value
Diluted-sample depletion, $\Delta D$	5.0 mg/L
Blank depletion, $\Delta B$	0.2 mg/L
BOD5 of undiluted raw sewage	240 mg/L (medium-strength domestic range)

Concept & Theory

Topic. EDTA hardness titration and the dilution/seed-corrected BOD test.

Key relations. $\text{Hardness (mg/L as CaCO}_3)=\dfrac{V_t\,N\,(50)(1000)}{V_s}$; $\text{BOD}_5=\dfrac{(D_1-D_2)-(B_1-B_2)}{P}$.

Why this works. EDTA's 1:1, charge-based complexation with Ca/Mg means titrant equivalents equal hardness equivalents directly, so a single conversion through CaCO₃'s equivalent weight gives the answer. The BOD dilution/seed formula isolates the sample's true oxygen demand by removing the dilution water's own background respiration (the blank term) before rescaling the small bottle fraction back up to 100% sewage strength (dividing by $P$) — without both corrections the reported BOD would be biased by whatever oxygen the blank consumed on its own and would still reflect the diluted, not the raw, concentration.

Common pitfall. Forgetting the equivalent-weight-of-50 conversion and reporting hardness in eq/L or mmol/L instead of mg/L as CaCO₃; omitting the blank correction (or applying it in the wrong direction) in the BOD calculation; dividing by the wrong dilution ratio (using the dilution WATER'S fraction, 98%, instead of the sample's own fraction, 2%).

Source. Davis & Cornwell, Introduction to Environmental Engineering, hardness/water-quality-analysis chapter; Metcalf & Eddy, Wastewater Engineering, ch. 3 (BOD test procedure and dilution/seed correction).

Question 2 of 5: Hardness Titration, the BOD Blank, and BOD of Raw Sewage

Question 2: Hardness Titration, the BOD Blank, and BOD of Raw Sewage (25 marks)

(a) Hardness by EDTA Titration

(b) Importance of the "Blank" in the Standard BOD Test

(c) BOD5 of the Undiluted Raw Sewage