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🔬 WQA Formula Sheet

Water Quality Analyst — Ontario Certification Exam Reference

📏 Unit Conversions🧪 Dilution & Standards💧 Alkalinity & Hardness☣️ CT Values⚖️ Langelier Index📊 QA/QC📋 Regulatory Limits

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Unit Conversions

mg/L ↔ µg/L

µg/L = mg/L × 1,000     |     mg/L = µg/L ÷ 1,000

Worked Example

A sample contains 0.045 mg of fluoride in 50 mL. Express the concentration in µg/L.

Concentration (mg/L) = 0.045 mg ÷ 0.050 L = 0.90 mg/L 0.90 mg/L × 1,000 = 900 µg/L

Answer: 900 µg/L

💡 Exam tip: 1 mg/L = 1 ppm (parts per million) in dilute aqueous solutions. 1 µg/L = 1 ppb. The WQA exam frequently asks you to convert between these — always check which unit the question asks for.

g/L ↔ mg/L

mg/L = g/L × 1,000     |     g/L = mg/L ÷ 1,000

Worked Example

A stock solution contains 5.85 g/L NaCl. What is this in mg/L?

5.85 g/L × 1,000 = 5,850 mg/L

Answer: 5,850 mg/L

💡 Exam tip: Reagent concentrations are often given in g/L; sample results are reported in mg/L. Know both directions cold.

Concentration from Mass & Volume

C (mg/L) = mass (mg) ÷ volume (L)

Variables / Reference Values

CConcentration (mg/L)

massMass of solute (mg)

volumeVolume of solution (L)

Worked Example

A 250 mL sample contains 0.125 mg of nitrate. What is the concentration in mg/L?

C = 0.125 mg ÷ 0.250 L = 0.50 mg/L

Answer: 0.50 mg/L

💡 Exam tip: Convert mL to L before dividing. This is the most fundamental calculation on the WQA exam.

Normality ↔ Molarity

N = M × n     |     M = N ÷ n

Variables / Reference Values

NNormality (eq/L)

MMolarity (mol/L)

nn-factor (equivalents per mole; for H₂SO₄ = 2, for NaOH = 1, for HCl = 1)

Worked Example

What is the normality of a 0.05 M H₂SO₄ solution?

N = 0.05 mol/L × 2 eq/mol = 0.10 N

Answer: 0.10 N

💡 Exam tip: For acid-base titrations: n-factor = number of H⁺ or OH⁻ ions exchanged. H₂SO₄ has n=2; HCl and NaOH have n=1.

Equivalent Weight & Normality

N = (mass in g/L) ÷ EW     |     EW = MW ÷ n

Variables / Reference Values

EWEquivalent weight (g/eq)

MWMolecular weight (g/mol)

nn-factor

Worked Example

What is the normality of a solution containing 4.9 g/L of H₂SO₄ (MW = 98 g/mol)?

EW = 98 ÷ 2 = 49 g/eq N = 4.9 g/L ÷ 49 g/eq = 0.10 N

Answer: 0.10 N

💡 Exam tip: EW of CaCO₃ = 100.09/2 = 50.05 g/eq. This is used constantly in hardness and alkalinity calculations.

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Dilution & Standard Preparation

Dilution Equation (C₁V₁ = C₂V₂)

C₁ × V₁ = C₂ × V₂

Variables / Reference Values

C₁Concentration of stock solution

V₁Volume of stock solution needed

C₂Desired concentration of working standard

V₂Final volume of working standard

Worked Example

How many mL of a 1,000 mg/L stock standard are needed to prepare 100 mL of a 5 mg/L working standard?

V₁ = (C₂ × V₂) ÷ C₁ = (5 mg/L × 100 mL) ÷ 1,000 mg/L = 500 ÷ 1,000 = 0.5 mL

Answer: 0.5 mL of stock + 99.5 mL diluent

💡 Exam tip: Units of C must match on both sides. Units of V must match on both sides. Rearrange to solve for the unknown: V₁ = (C₂ × V₂) ÷ C₁.

Serial Dilution Factor

DF = V_sample ÷ V_total     |     C_final = C_initial × DF₁ × DF₂ × …

Variables / Reference Values

DFDilution factor for one step (dimensionless, 0–1)

V_sampleVolume of sample taken (mL)

V_totalTotal volume after dilution (mL)

Worked Example

A sample is diluted 1:10, then that dilution is diluted 1:5. What is the overall dilution factor?

Overall DF = (1/10) × (1/5) = 1/50 = 0.02

Answer: 1:50 overall dilution (DF = 0.02)

💡 Exam tip: A 1:10 dilution has DF = 0.1. Two sequential 1:10 dilutions give an overall DF = 0.01 (1:100). Always multiply DFs for serial dilutions.

Back-Calculation to Original Concentration

C_original = C_measured ÷ DF

Variables / Reference Values

C_originalTrue concentration in original sample

C_measuredConcentration measured in diluted sample

DFOverall dilution factor applied

Worked Example

A sample diluted 1:20 gives a reading of 3.5 mg/L. What is the concentration in the original sample?

C_original = 3.5 mg/L ÷ (1/20) = 3.5 × 20 = 70 mg/L

Answer: 70 mg/L

💡 Exam tip: If a 1:100 diluted sample reads 0.25 mg/L, the original is 0.25 ÷ 0.01 = 25 mg/L. Always divide by the dilution factor, not multiply.

Percent Recovery (Spike Recovery)

% Recovery = (C_spiked − C_unspiked) ÷ C_spike_added × 100

Variables / Reference Values

C_spikedMeasured concentration in spiked sample (mg/L)

C_unspikedMeasured concentration in unspiked sample (mg/L)

C_spike_addedKnown concentration of spike added (mg/L)

Worked Example

An unspiked sample reads 1.2 mg/L. After adding a 2.0 mg/L spike, the spiked sample reads 3.0 mg/L. What is the % recovery?

% Recovery = (3.0 − 1.2) ÷ 2.0 × 100 = 1.8 ÷ 2.0 × 100 = 90%

Answer: 90% — within acceptable range (80–120%)

💡 Exam tip: Acceptable range is typically 80–120% for most WQA parameters. Values outside this range indicate matrix interference, contamination, or instrument drift.

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Alkalinity & Hardness

Total Alkalinity (Titration)

Alkalinity (mg/L as CaCO₃) = (V_acid × N_acid × 50,000) ÷ V_sample

Variables / Reference Values

V_acidVolume of acid titrant used (mL)

N_acidNormality of acid titrant (eq/L)

50,000Conversion factor: EW of CaCO₃ (50.05) × 1,000 mL/L

V_sampleVolume of sample titrated (mL)

Worked Example

12.5 mL of 0.02 N H₂SO₄ is used to titrate 100 mL of sample. What is the total alkalinity?

Alkalinity = (12.5 × 0.02 × 50,000) ÷ 100 = 12,500 ÷ 100 = 125 mg/L as CaCO₃

Answer: 125 mg/L as CaCO₃

💡 Exam tip: The factor 50,000 = EW of CaCO₃ × 1,000. Memorize this — it appears on almost every WQA exam. Phenolphthalein alkalinity uses the P endpoint (pH 8.3); total alkalinity uses the M endpoint (pH 4.5).

Hardness (EDTA Titration)

Hardness (mg/L as CaCO₃) = (V_EDTA × M_EDTA × 100,090) ÷ V_sample

Variables / Reference Values

V_EDTAVolume of EDTA titrant used (mL)

M_EDTAMolarity of EDTA (mol/L)

100,090MW of CaCO₃ (g/mol) × 1,000 mL/L

V_sampleVolume of sample titrated (mL)

Worked Example

8.5 mL of 0.01 M EDTA is used to titrate 50 mL of sample. What is the total hardness?

Hardness = (8.5 × 0.01 × 100,090) ÷ 50 = 8,507.65 ÷ 50 ≈ 170 mg/L as CaCO₃

Answer: 170 mg/L as CaCO₃

💡 Exam tip: EDTA forms 1:1 complexes with Ca²⁺ and Mg²⁺. Eriochrome Black T indicator turns blue at the endpoint. Use pH 10 buffer. Magnesium hardness = Total hardness − Calcium hardness.

Hardness Conversion: CaCO₃ → Ion

C_ion (mg/L) = Hardness (mg/L as CaCO₃) × (MW_ion ÷ (n × EW_CaCO₃))

Variables / Reference Values

MW_ionAtomic/molecular weight of ion (Ca²⁺ = 40.08, Mg²⁺ = 24.31)

nValence of ion (Ca²⁺ = 2, Mg²⁺ = 2)

EW_CaCO₃Equivalent weight of CaCO₃ = 50.05 g/eq

Worked Example

Total hardness = 200 mg/L as CaCO₃; calcium hardness = 120 mg/L as CaCO₃. What is the Mg²⁺ concentration?

Mg hardness = 200 − 120 = 80 mg/L as CaCO₃ Mg²⁺ = 80 × 0.2431 = 19.4 mg/L

Answer: 19.4 mg/L as Mg²⁺

💡 Exam tip: Shortcut: Ca²⁺ (mg/L) = Ca hardness (mg/L as CaCO₃) × 0.4008. Mg²⁺ (mg/L) = Mg hardness (mg/L as CaCO₃) × 0.2431.

meq/L Conversion

meq/L = (mg/L as CaCO₃) ÷ 50     |     mg/L as CaCO₃ = meq/L × 50

Worked Example

A sample has a total alkalinity of 150 mg/L as CaCO₃. Express this in meq/L.

meq/L = 150 ÷ 50 = 3.0 meq/L

Answer: 3.0 meq/L

💡 Exam tip: The equivalent weight of CaCO₃ is 50.05 g/eq ≈ 50. So 1 meq/L = 50 mg/L as CaCO₃. This conversion is used in ion balance and corrosion index calculations.

☣️

CT Values & Disinfection

CT Value

CT = C × T

Units: mg·min/L

Variables / Reference Values

CDisinfectant residual concentration (mg/L) — measured at the outlet of the contact chamber

TContact time (min) — use T₁₀ (time for 10% of flow to pass through)

Worked Example

A chlorine contact chamber has a residual of 1.5 mg/L and a T₁₀ of 45 minutes. What is the CT?

CT = 1.5 mg/L × 45 min = 67.5 mg·min/L

Answer: 67.5 mg·min/L

💡 Exam tip: T₁₀ = Baffling factor × HRT. Baffling factors: excellent = 0.7, good = 0.5, average = 0.3, poor = 0.1. Always use T₁₀, not HRT, in CT calculations.

CT Required for Giardia (Free Chlorine, 10°C)

CT_required = 165 mg·min/L (for 4-log / 99.99% inactivation at 10°C, pH 7–8)

💡 Exam tip: Ontario's Procedure for Disinfection requires 4-log (99.99%) Giardia inactivation. CT requirements increase at lower temperatures and higher pH. At 5°C the CT requirement is approximately 235 mg·min/L. UV can also achieve Giardia credit at 40 mJ/cm² for 4-log.

CT Required for Viruses (Free Chlorine, 10°C)

CT_required = 6 mg·min/L (for 4-log / 99.99% inactivation at 10°C, pH 6–9)

💡 Exam tip: Viruses require much less CT than Giardia because free chlorine is highly effective against viruses. UV at 40 mJ/cm² also achieves 4-log virus inactivation credit in Ontario.

Log Inactivation Credit

Log inactivation = log₁₀(N₀ ÷ N)

Variables / Reference Values

N₀Initial organism concentration

NFinal organism concentration after disinfection

Worked Example

A system achieves 2-log removal through filtration and 2-log inactivation through disinfection. What is the total log credit for Giardia?

Total log credit = 2 + 2 = 4-log (99.99% inactivation)

Answer: 4-log — meets Ontario's minimum requirement

💡 Exam tip: 1-log = 90% removal; 2-log = 99%; 3-log = 99.9%; 4-log = 99.99%. Ontario requires 4-log for both Giardia and viruses from combined treatment and disinfection.

Chlorine Dose, Demand & Residual

Dose = Demand + Residual

Variables / Reference Values

DoseAmount of chlorine added (mg/L)

DemandChlorine consumed by reactions with organics, ammonia, metals (mg/L)

ResidualChlorine remaining after demand is satisfied (mg/L)

Worked Example

A water system adds 3.5 mg/L of chlorine. The measured residual is 0.8 mg/L. What is the chlorine demand?

Demand = Dose − Residual = 3.5 − 0.8 = 2.7 mg/L

Answer: 2.7 mg/L chlorine demand

💡 Exam tip: O. Reg. 170/03 requires a minimum free chlorine residual of 0.05 mg/L throughout the distribution system. Most systems target 0.2 mg/L at the extremities.

UV Dose

UV Dose (mJ/cm²) = UV Intensity (mW/cm²) × Exposure Time (s)

Worked Example

A UV reactor delivers 8 mW/cm² at the required flow rate. How many seconds of exposure are needed for a 40 mJ/cm² dose?

Time = 40 mJ/cm² ÷ 8 mW/cm² = 5 seconds

Answer: 5 seconds

💡 Exam tip: Ontario requires 40 mJ/cm² for 4-log Giardia and 4-log virus inactivation credit. UV dose must be validated at the lowest transmittance (UVT) and highest flow rate expected in operation.

⚖️

Langelier Saturation Index

Langelier Saturation Index (LSI)

LSI = pH − pHs

Variables / Reference Values

pHMeasured pH of the water

pHspH at which the water is saturated with CaCO₃ (calculated)

Worked Example

A water sample has pH 7.8 and a calculated pHs of 8.4. What is the LSI and what does it indicate?

LSI = 7.8 − 8.4 = −0.6

Answer: LSI = −0.6 — water is corrosive (will dissolve CaCO₃ scale and attack pipe surfaces)

💡 Exam tip: LSI > 0: scale-forming (CaCO₃ will precipitate). LSI < 0: corrosive (CaCO₃ will dissolve). LSI = 0: balanced. Target LSI of 0 to +0.5 for corrosion control in distribution systems.

pHs Calculation

pHs = (9.3 + A + B) − (C + D)

Variables / Reference Values

ATDS factor: log₁₀(TDS) − 1 (or use lookup table)

BTemperature factor (from lookup table; decreases with temperature)

CCalcium hardness factor: log₁₀(Ca hardness as CaCO₃)

DAlkalinity factor: log₁₀(Total alkalinity as CaCO₃)

Worked Example

Given: pHs = 8.1, measured pH = 7.6. What is the LSI?

LSI = 7.6 − 8.1 = −0.5

Answer: LSI = −0.5 (slightly corrosive)

💡 Exam tip: The WQA exam typically provides pHs or the lookup table values. Know the formula structure and that higher Ca hardness and alkalinity lower pHs (making water more scale-forming).

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QA/QC Calculations

Relative Percent Difference (RPD)

RPD (%) = |C₁ − C₂| ÷ ((C₁ + C₂) ÷ 2) × 100

Variables / Reference Values

C₁Concentration in first replicate

C₂Concentration in second replicate

Worked Example

A duplicate pair gives results of 4.8 mg/L and 5.2 mg/L. What is the RPD?

RPD = |4.8 − 5.2| ÷ ((4.8 + 5.2) ÷ 2) × 100 = 0.4 ÷ 5.0 × 100 = 8%

Answer: 8% — acceptable (< 20%)

💡 Exam tip: RPD measures precision between duplicates. Acceptable RPD is typically ≤ 20% for most parameters. High RPD indicates poor precision — check homogeneity, analyst technique, or instrument stability.

Method Detection Limit (MDL)

MDL = t_(n-1, 99%) × s

Variables / Reference Values

t_(n-1, 99%)Student's t-value at 99% confidence for (n−1) degrees of freedom

sStandard deviation of n replicate measurements of a low-level spike

nNumber of replicates (minimum 7 per EPA MDL procedure)

Worked Example

Seven replicates of a low-level spike give a standard deviation of 0.012 mg/L. What is the MDL? (t = 3.143)

MDL = 3.143 × 0.012 = 0.038 mg/L

Answer: MDL = 0.038 mg/L

💡 Exam tip: MDL is the lowest concentration that can be detected with 99% confidence. For n=7 replicates, t = 3.143. The MDL must be ≤ 1/3 of the regulatory limit for the method to be valid.

Z-Score (Proficiency Testing)

Z = (x − X) ÷ σ

Variables / Reference Values

xLab's reported result

XAssigned (true) value from PT provider

σStandard deviation of the PT program

Worked Example

A lab reports 1.85 mg/L for a PT sample with an assigned value of 2.00 mg/L and σ = 0.10 mg/L. What is the Z-score?

Z = (1.85 − 2.00) ÷ 0.10 = −0.15 ÷ 0.10 = −1.5

Answer: Z = −1.5 — satisfactory (|Z| ≤ 2)

💡 Exam tip: |Z| ≤ 2: satisfactory. 2 < |Z| ≤ 3: questionable (warning). |Z| > 3: unsatisfactory (action required). Proficiency testing is mandatory for accredited labs under O. Reg. 252/05.

Coefficient of Variation (CV)

CV (%) = (s ÷ x̄) × 100

Variables / Reference Values

sStandard deviation of replicate measurements

x̄Mean (average) of replicate measurements

Worked Example

Five replicates of a chlorine standard give: 1.02, 0.98, 1.05, 1.00, 0.95 mg/L. Mean = 1.00, s = 0.037. What is the CV?

CV = (0.037 ÷ 1.00) × 100 = 3.7%

Answer: CV = 3.7% — excellent precision

💡 Exam tip: CV expresses precision as a percentage of the mean. Lower CV = better precision. CV < 10% is generally acceptable for routine water analysis; < 5% for critical parameters.

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Regulatory Limits (Ontario)

Key MACs — O. Reg. 169/03

Maximum Acceptable Concentrations (MACs) for drinking water parameters

Variables / Reference Values

Nitrate (NO₃-N)MAC = 10 mg/L as N (= 44.3 mg/L as NO₃)

Nitrite (NO₂-N)MAC = 1.0 mg/L as N

FluorideMAC = 1.5 mg/L; AO = 1.0 mg/L (optimal 0.7 mg/L)

LeadMAC = 0.010 mg/L (10 µg/L)

ArsenicMAC = 0.010 mg/L (10 µg/L)

Turbidity (treated)MAC = 1.0 NTU; operational trigger = 0.3 NTU (conventional filtration)

E. coliMAC = 0 CFU/100 mL (zero tolerance)

Total ColiformsMAC = 0 CFU/100 mL in treated water

THMs (total)MAC = 0.100 mg/L (100 µg/L)

HAAs (total)MAC = 0.080 mg/L (80 µg/L)

💡 Exam tip: These are the most frequently tested regulatory values on the WQA exam. MACs are health-based limits; Aesthetic Objectives (AOs) are non-health-based guidelines.

Chlorine Residual Requirements — O. Reg. 170/03

Minimum free Cl₂ residual = 0.05 mg/L at all points in distribution system

💡 Exam tip: The 0.05 mg/L minimum is the regulatory floor — not a target. Most systems aim for 0.2 mg/L at extremities. Maximum residual is not specified in O. Reg. 170/03 but Health Canada's aesthetic objective for chlorine is 0.6 mg/L (taste/odour threshold).

Adverse Test Result Notification

Notify MOH: immediately and no later than 24 hours after becoming aware

💡 Exam tip: Under O. Reg. 170/03 s.18, an adverse test result (E. coli, total coliforms, turbidity exceedance, or MAC exceedance) requires immediate notification to the local Medical Officer of Health. The 24-hour clock starts when the owner becomes aware — not when the lab reports.

Record Retention

Operational records must be retained for a minimum of 10 years (O. Reg. 170/03)

💡 Exam tip: This includes test results, maintenance logs, calibration records, and operator logs. The 10-year requirement applies to all records. Some records (e.g., design drawings) must be kept indefinitely.