19.3 Fluoride Chemical Feed Calculations & Solution Dilutions

Chemistry of Fluoride Dosing and Available Fluoride Ion (AFI)

Community water fluoridation is the controlled addition of a fluoride compound to drinking water to reduce tooth decay in the consumer population. Unlike chlorine or coagulants, where chemical addition is adjusted to satisfy an operational demand, fluoride is dosed to maintain a strict, steady concentration throughout the distribution network.

                +-------------------------------------------------------+
                |        Raw Water Natural Fluoride (mg/L)              |
                +-------------------------------------------------------+
                                           |
                                           v  Target (0.7 mg/L) - Raw
                +-------------------------------------------------------+
                |              Dose Needed (mg/L)                       |
                +-------------------------------------------------------+
                                           |
                                           v  × Flow (MGD) × 8.34 lb/gal
                +-------------------------------------------------------+
                |           Pure Fluoride Required (lb/day)             |
                +-------------------------------------------------------+
                                           |
                                           v  ÷ (AFI × Purity Decimal)
                +-------------------------------------------------------+
                |         Commercial Fluoride Chemical (lb/day)         |
                +-------------------------------------------------------+

Regulatory Concentration Standards

The U.S. Environmental Protection Agency (EPA) and U.S. Public Health Service (PHS) enforce specific drinking water fluoride benchmarks:

• Recommended Optimum Level ($0.7\ mg/L$): In 2015, the U.S. Department of Health and Human Services (HHS) updated the national recommendation to a single uniform concentration of $0.7\ mg/L$ (replacing the historical $0.7\text{ to }1.2\ mg/L$ climate-dependent range). This level provides optimal cavity prevention while minimizing dental fluorosis.
• Secondary Maximum Contaminant Level ($SMCL = 2.0\ mg/L$): A non-enforceable federal aesthetic guideline. Concentrations exceeding $2.0\ mg/L$ can cause cosmetic dental fluorosis (brown staining and enamel pitting in children under 9 years of age). Water systems exceeding $2.0\ mg/L$ must issue a public notice.
• Maximum Contaminant Level ($MCL = 4.0\ mg/L$): An enforceable National Primary Drinking Water Regulation health standard. Chronic exposure above $4.0\ mg/L$ causes crippling skeletal fluorosis (bone density degradation and joint stiffness).

Concept of Available Fluoride Ion (AFI)

Commercial fluoride chemicals are chemical compounds—salts or complex acids—composed of fluoride combined with other elements like sodium, silicon, and hydrogen. They do not consist of pure elemental fluoride ($F^-$). The fraction of pure fluoride ion contained within a chemical compound is dictated by its stoichiometry and is termed the Available Fluoride Ion (AFI) fraction:

$$\text{AFI} = \frac{\text{Total Molecular Weight of Fluorine in Compound}}{\text{Total Molecular Weight of Chemical Compound}}$$

The Three Approved Drinking Water Fluoride Chemicals

Water utilities are authorized to use three standardized chemical compounds for fluoridation:

1. Sodium Fluoride ($\text{NaF}$):

   • Physical State: White, odorless crystalline powder or fine granules; widely used in small systems and saturator tanks.
   • Molecular Weight ($MW$): Sodium ($23.0$) + Fluorine ($19.0$) = $42.0\ g/mol$.
   • Available Fluoride Ion (AFI): $$\text{AFI} = \frac{19.0}{42.0} = 0.45238 \approx \mathbf{45.3\% \ (0.453)}$$
   • Commercial Purity: Typically $98.0\%$ pure ($0.98$ decimal).
   • Combined Active Fluoride Fraction: $0.453 \times 0.98 = \mathbf{0.444\ (44.4\%)}$.

2. Sodium Fluorosilicate / Sodium Silicofluoride ($\text{Na}_2\text{SiF}_6$):

   • Physical State: White, free-flowing crystalline powder; fed through dry volumetric or gravimetric feeders in medium to large plants.
   • Molecular Weight ($MW$): $(2 \times 23.0) + 28.1 + (6 \times 19.0) = 46.0 + 28.1 + 114.0 = 188.1\ g/mol$.
   • Available Fluoride Ion (AFI): $$\text{AFI} = \frac{6 \times 19.0}{188.1} = \frac{114.0}{188.1} = 0.60606 \approx \mathbf{60.7\% \ (0.607)}$$
   • Commercial Purity: Typically $98.5\%$ pure ($0.985$ decimal).
   • Combined Active Fluoride Fraction: $0.607 \times 0.985 = \mathbf{0.598\ (59.8\%)}$.

3. Fluorosilicic Acid / Hydrofluorosilicic Acid ($\text{H}_2\text{SiF}_6$):

   • Physical State: Transparent, straw-colored, fuming, highly corrosive liquid; most common chemical in automated facilities.
   • Molecular Weight ($MW$): $(2 \times 1.0) + 28.1 + (6 \times 19.0) = 2.0 + 28.1 + 114.0 = 144.1\ g/mol$.
   • Available Fluoride Ion (AFI): $$\text{AFI} = \frac{6 \times 19.0}{144.1} = \frac{114.0}{144.1} = 0.79112 \approx \mathbf{79.2\% \ (0.792)}$$
   • Commercial Solution Strength: Typically delivered as an aqueous solution between $23.0\%$ and $25.0\%$ acid ($0.23\text{ to }0.25$ decimal).
   • Specific Gravity: Typically $1.22$ ($10.17\ lb/gal$).
   • Active Fluoride per Gallon ($24\%$ acid): $$\text{Active }\text{F}^- = 1.22 \times 8.34\ lb/gal \times 0.24 \times 0.792 = \mathbf{1.933\ lb\,\text{F}^-/gal}$$

---

The Universal Fluoride Feed Formula

Calculating fluoride chemical feed requirements requires a two-step sequence: determining the incremental fluoride dose needed, and adjusting for chemical compound purity and AFI.

Step 1: Incremental Dosage Determination

Virtually all raw surface water and groundwater sources contain natural background concentrations of fluoride dissolved from mineral strata. The chemical feed system must only deliver the difference between target finished concentration and natural background concentration:

$$\text{Dose Needed }(mg/L) = \text{Target Concentration }(0.70\ mg/L) - \text{Raw Water Background }(mg/L)$$

Exam Warning: If an exam problem states that raw water already contains $0.20\ mg/L$ fluoride and the target is $0.70\ mg/L$, the chemical feeder must be set to deliver only $0.50\ mg/L$ ($0.70 - 0.20 = 0.50\ mg/L$). Feeding the full $0.70\ mg/L$ would overfeed the system to $0.90\ mg/L$.

Step 2: Applying the Universal Fluoride Equation

To calculate commercial fluoride product feed rate in pounds per day ($lb/day$):

$$\text{Commercial Fluoride Feed }(lb/day) = \frac{\text{Flow }(MGD) \times \text{Dose Needed }(mg/L) \times 8.34\ lb/gal}{\text{AFI} \times \text{Purity Decimal}}$$

For liquid fluorosilicic acid ($\text{H}_2\text{SiF}_6$), dividing commercial pounds per day by solution density ($SG \times 8.34$) yields the liquid feed rate in gallons per day ($gpd$):

$$\text{Liquid Acid Feed }(gpd) = \frac{\text{Commercial Fluoride Feed }(lb/day)}{\text{Acid Solution Density }(lb/gal)}$$ $$\text{Liquid Acid Feed }(gpd) = \frac{\text{Flow }(MGD) \times \text{Dose Needed }(mg/L) \times 8.34\ lb/gal}{\text{AFI} \times \text{Acid Concentration Decimal} \times (SG \times 8.34\ lb/gal)}$$ $$\mathbf{\text{Liquid Acid Feed }(gpd) = \frac{\text{Flow }(MGD) \times \text{Dose Needed }(mg/L)}{\text{AFI} \times \text{Acid Concentration Decimal} \times SG}}$$

---

Sodium Fluoride Saturator Calculations and Hydraulics

Small to medium-sized water utilities frequently feed sodium fluoride using an upflow or downflow saturator. The saturator utilizes a unique physical property of sodium fluoride: across the entire range of normal drinking water temperatures ($32^\circ\text{F}\text{ to }100^\circ\text{F} / 0^\circ\text{C}\text{ to }38^\circ\text{C}$), the solubility of $\text{NaF}$ in water remains remarkably constant at $4.0\%$ by weight ($4.0\text{ grams }\text{NaF}$ per $100\ mL$ water, or $40,000\ mg/L\,\text{NaF}$).

Constant Available Fluoride Concentration in Saturated Solution

Since sodium fluoride maintains a constant $40,000\ mg/L\,\text{NaF}$ concentration and has an AFI of $45.3\%$ ($0.4524$):

$$\text{Fluoride Concentration in Saturated Solution} = 40,000\ mg/L\,\text{NaF} \times 0.4524\,\text{AFI} = \mathbf{18,096\ mg/L\,\text{F}^-}$$

On state operator certification exams and across water utility industry standards, this concentration is universally rounded to $18,000\ mg/L\,\text{F}^-$.

Saturator Feed Rate Equation

Because the saturator solution always delivers exactly $18,000\ mg/L\,\text{F}^-$, operators calculate the required volumetric delivery rate directly:

$$\text{Saturator Feed }(gal/day) = \frac{\text{Flow }(MGD) \times \text{Dose Needed }(mg/L) \times 1,000,000\ gal/MG}{18,000\ mg/L}$$ $$\mathbf{\text{Saturator Feed }(gal/day) = \frac{\text{Flow }(gpd) \times \text{Dose Needed }(mg/L)}{18,000\ mg/L}}$$

The Golden Rule of Saturator Sizing

To dose $1.0\ mg/L$ of fluoride into $1,000\text{ gallons}$ of water requires:

$$\text{Saturator Solution} = \frac{1,000\ gal \times 1.0\ mg/L}{18,000\ mg/L} = \frac{1}{18}\text{ gallon} = 0.05556\text{ gallons} \approx 7.1\text{ fluid ounces}$$

To dose $1.0\ mg/L$ into $1.0\text{ million gallons } (1.0\ MGD)$ of water requires exactly $55.56\ gallons$ of saturated solution. To deliver the recommended optimum dose of $0.70\ mg/L$ into $1.0\ MGD$ requires $38.89\ gallons$ of solution.

Operating Mandate for Saturator Water: Saturator makeup water must always be softened if raw water total hardness exceeds $50\text{ to }75\ mg/L\text{ as }\text{CaCO}_3$. In hard water, calcium and magnesium ions react with dissolved fluoride to form calcium fluoride ($\text{CaF}_2$), an insoluble precipitate that coats bed granules, cements the sand bed into a solid block, and clogs suction piping.

---

Solution Dilution Mathematics (The Two-Normal Equation)

In many treatment applications—including preparing liquid polymer batches, creating working chemical day tanks, and diluting concentrated stock bleach or acid—operators must prepare a weaker target concentration from a concentrated chemical stock.

The Principle of Mass Conservation

When clean water is added to a chemical solution, the total volume increases while the chemical concentration decreases; however, the absolute mass of active chemical molecules remains unchanged. This mass conservation is expressed by the Two-Normal / Concentration-Volume Equation:

$$C_1 \times V_1 = C_2 \times V_2$$

Where:

• $C_1$: Concentration of the original concentrated stock chemical (in $\%$ or $mg/L$).
• $V_1$: Volume of concentrated stock chemical required.
• $C_2$: Target concentration of the final diluted working solution (in $\%$ or $mg/L$).
• $V_2$: Total target volume of the final diluted working solution.

Rearranging to solve for the volume of concentrated chemical ($V_1$):

$$V_1 = \frac{C_2 \times V_2}{C_1}$$

Volume of Dilution Water Required

After calculating the volume of concentrated chemical ($V_1$), the required volume of make-up water is the difference between total final batch volume and stock chemical volume:

$$\text{Make-up Dilution Water Volume} = V_2 - V_1$$

Critical Chemical Safety Rule: Always add concentrated acid or chemical to water—never add water to concentrated chemical (AAA: Always Add Acid). Adding water to concentrated chemical triggers an intense localized exothermic reaction, causing violent boiling and acid splatter.

---

Step-by-Step Worked Practice Problems

Example 1: Hydrofluorosilicic Acid ($H_2SiF_6$) Feed Calculation

Problem Statement: A regional water plant treats a steady flow of $4.8\text{ MGD}$. Laboratory testing shows the raw water has a natural background fluoride concentration of $0.15\text{ mg/L}$. The plant must achieve the recommended finished water target of $0.70\text{ mg/L}$. The utility feeds commercial fluorosilicic acid at $24.0\%$ concentration by weight with a specific gravity of $1.22$ and an Available Fluoride Ion (AFI) fraction of $0.792$. Calculate:

1. The incremental fluoride dose needed in $mg/L$.
2. The mass of pure fluoride ion required in pounds per day ($lb/day$).
3. The commercial fluorosilicic acid required in pounds per day ($lb/day$).
4. The liquid acid feed rate in gallons per day ($gpd$) and milliliters per minute ($mL/min$).

Solution Procedure:

• Step 1: Calculate Incremental Fluoride Dose Needed

  $$\text{Dose Needed} = \text{Target} - \text{Raw} = 0.70\ mg/L - 0.15\ mg/L = 0.55\ mg/L$$

• Step 2: Calculate Pure Fluoride Ion Required ($lb/day$)

  $$\text{Pure }\text{F}^- (lb/day) = 4.8\ MGD \times 0.55\ mg/L \times 8.34\ lb/gal = 22.0176\ lb\,\text{F}^-/day$$

• Step 3: Calculate Commercial Product Feed Rate ($lb/day$)
  Adjust for both AFI ($0.792$) and acid concentration ($0.24$):

  $$\text{Combined Active Fraction} = 0.792 \times 0.24 = 0.19008$$ $$\text{Commercial Acid Feed }(lb/day) = \frac{22.0176\ lb/day}{0.19008} = 115.833\dots\ lb/day \approx 115.8\ lb/day$$

• Step 4: Convert to Volumetric Liquid Feed Rate

  $$\text{Acid Density} = 1.22 \times 8.34\ lb/gal = 10.1748\ lb/gal$$ $$\text{Liquid Feed Rate }(gpd) = \frac{115.833\ lb/day}{10.1748\ lb/gal} = 11.384\ gpd \approx 11.38\ gpd$$ $$\text{Delivery }(mL/min) = 11.384\ gpd \times 2.62847 = 29.92\ mL/min \approx 29.9\ mL/min$$

Example 2: Sodium Fluoride Saturator Daily Feed

Problem Statement: An operator operates an upflow sodium fluoride saturator treating a well supply flowing at $450,000\text{ gallons per day}$ ($0.45\text{ MGD}$). The raw groundwater contains $0.10\text{ mg/L}$ natural fluoride, and the target finished water concentration is $0.70\text{ mg/L}$. What is the required saturator solution feed rate in gallons per day ($gpd$) and milliliters per minute ($mL/min$)?

Solution Procedure:

• Step 1: Calculate Dose Needed

  $$\text{Dose Needed} = 0.70\ mg/L - 0.10\ mg/L = 0.60\ mg/L$$

• Step 2: Solve for Daily Saturator Feed Volume

  $$\text{Saturator Feed }(gpd) = \frac{\text{Daily Flow }(gpd) \times \text{Dose }(mg/L)}{18,000\ mg/L}$$ $$\text{Saturator Feed} = \frac{450,000\ gpd \times 0.60\ mg/L}{18,000\ mg/L} = \frac{270,000}{18,000} = \mathbf{15.0\ gpd}$$

• Step 3: Convert to Delivery Rate in $mL/min$

  $$\text{Feed Rate }(mL/min) = 15.0\ gpd \times 2.62847 = 39.427\ mL/min \approx 39.4\ mL/min$$

Example 3: Chemical Solution Dilution ($C_1 \times V_1 = C_2 \times V_2$)

Problem Statement: An operator must prepare $120\text{ gallons}$ of a $1.5\%$ sodium hypochlorite solution in a chemical day tank by diluting a $10.0\%$ bulk commercial bleach stock solution with potable water. How many gallons of the $10.0\%$ stock bleach and how many gallons of dilution water are required?

Solution Procedure:

• Step 1: Identify Given Variables for the Dilution Equation

  • $C_1 = 10.0\%$ (stock concentration)
  • $V_1 = ?$
  • $C_2 = 1.5\%$ (target concentration)
  • $V_2 = 120.0\text{ gallons}$ (total batch volume)

• Step 2: Solve for Stock Solution Volume ($V_1$)

  $$C_1 \times V_1 = C_2 \times V_2 \implies V_1 = \frac{C_2 \times V_2}{C_1}$$ $$V_1 = \frac{1.5\% \times 120.0\ gal}{10.0\%} = \frac{180.0}{10.0} = \mathbf{18.0\ gallons\text{ of 10.0\% stock bleach}}$$

• Step 3: Calculate Required Volume of Dilution Water

  $$\text{Water Volume} = V_2 - V_1 = 120.0\ gal - 18.0\ gal = \mathbf{102.0\ gallons\text{ of water}}$$

---

Reference Tables: Fluoride Constants, Standards, and Saturator Feeding

Table 19.3.1: Physical, Chemical, and Handling Parameters of Fluoride Compounds

Table 19.3.2: Drinking Water Fluoride Concentrations and Regulatory Thresholds

Table 19.3.3: Sodium Fluoride Saturator Daily Delivery Rates ($gpd$ of Saturated Solution)