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Math problem set Grade 11 Math

Ospringe's Two Sources: pH on a Logarithmic Scale

A grade 11 problem set converting pH to hydrogen ion concentration for two fictional sources, blending them, and sizing the soda ash dose that follows.

Objectives

  1. Students will convert between pH and hydrogen ion concentration using logarithms and powers of ten.
  2. Students will explain why two waters cannot be blended by averaging their pH values.
  3. Students will compute a chemical feed rate in pounds per day from a dose and a plant flow.

Vocabulary

pH
pH is a number that tells whether water is acidic, like lemon juice, or basic, like baking soda mixed in water. Pure water sits in the middle of the scale at a pH of 7.

Also: hydrogen ion concentration, logarithm, soda ash, dose

Teacher overview

The pH scale reports the concentration of hydrogen ions as a negative base-ten logarithm, so a difference of one pH unit is a factor of ten in concentration rather than one step along a ruler. This problem set follows the fictional town of Ospringe, which blends water from a slightly acidic well with water from a brook intake and then has to decide how much soda ash to feed to bring the blend up to the range the secondary drinking water standard sets, which is a pH of 6.5 to 8.5. The misconception the set is built to correct is the habit of averaging two pH readings: students who average 6.1 and 7.8 report 6.95, while a simplified model that averages hydrogen ion concentrations puts the blend near 6.39, low enough to fall outside the standard. In real water, alkalinity and hydroxide ions also affect the result, so plants confirm a blend's pH by measurement or bench test.

Ospringe's Two Sources: pH on a Logarithmic Scale — Grade 11 Math
Name: ____________________________ Date: ______________

Problems

The town of Ospringe, its well, its brook intake, and its bench-test result are all fictional. Use these values throughout.

  • Ospringe Well No. 4 delivers water at pH 6.1.
  • The Ombersley Brook intake delivers water at pH 7.8.
  • The plant blends equal volumes of the two waters.
  • A bench test on this blend showed that each 1.0 mg/L of soda ash raises the pH by 0.09 units.
  • The plant treats 2.4 MGD.
  • Loading in pounds per day equals dose in mg/L times flow in MGD times 8.34.
  • The secondary standard for drinking water pH runs from 6.5 to 8.5.
  • Report concentrations to three significant figures and pH values to two decimal places.
  1. Find the hydrogen ion concentration of the well water in moles per liter.
  2. Find the hydrogen ion concentration of the brook water in moles per liter.
  3. How many times higher is the hydrogen ion concentration of the well water than that of the brook water?
  4. As a simplified model, assume the blended hydrogen ion concentration is the average of the two concentrations. Find it.
  5. Find the pH of the blend.
  6. Does the blend fall inside the secondary standard range of 6.5 to 8.5? Answer yes or no.
  7. The operator wants a finished pH of 7.50. Starting from the blend pH in problem 5, by what factor must the hydrogen ion concentration fall?
  8. Using the bench-test result, find the soda ash dose in mg/L that raises the blend from its pH in problem 5 to a pH of 7.50. Round to two decimal places.
  9. Using the dose from problem 8 and a flow of 2.4 MGD, find the soda ash feed rate in pounds per day. Round to one decimal place.
  10. The plant's field meter reads pH to the nearest 0.01 unit. Explain why reporting the blend pH from problem 5 as 6.3924 would misstate what the plant actually knows.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: 7.94 x 10^-7 mol/L.
    The concentration is 10 raised to the power of negative pH. Here 10^-6.1 = 10^0.9 x 10^-7 = 7.943 x 10^-7 moles per liter.
  2. Answer: 1.58 x 10^-8 mol/L.
    10^-7.8 = 10^0.2 x 10^-8 = 1.585 x 10^-8 moles per liter.
  3. Answer: 50.1 times.
    Dividing gives 7.943 x 10^-7 divided by 1.585 x 10^-8 = 50.12. The same result comes from the exponents directly, since 10^(7.8 - 6.1) = 10^1.7 = 50.1.
  4. Answer: 4.05 x 10^-7 mol/L.
    Write both values with the same exponent: 7.943 x 10^-7 and 0.1585 x 10^-7. Their sum is 8.1015 x 10^-7, and half of that is 4.051 x 10^-7 moles per liter.
  5. Answer: 6.39. pH is the negative base-ten logarithm of the concentration. The logarithm of 4.051 x 10^-7 is 0.6076 - 7 = -6.3924, so the pH is 6.39. Averaging the two original pH values would give 6.95, which under this model is more than half a unit too high, because the acidic source contributes fifty times as many hydrogen ions as the other source.
  6. Answer: No.
    A pH of 6.39 sits below the lower end of the 6.5 to 8.5 range, so the blend does not meet the standard without treatment.
  7. Answer: 12.9 times.
    The pH must rise by 7.50 - 6.39 = 1.11 units, and each unit is a factor of ten, so the concentration must fall by 10^1.11 = 12.88, which rounds to 12.9 times.
  8. Answer: 12.33 mg/L.
    The pH gap is 1.11 units, and each 1.0 mg/L of soda ash supplies 0.09 units, so the dose is 1.11 divided by 0.09 = 12.33 mg/L.
  9. Answer: 246.8 lb/day.
    Loading equals 12.33 mg/L x 2.4 MGD x 8.34 = 246.80 pounds per day. Working the multiplication in order, 12.33 x 2.4 = 29.592, and 29.592 x 8.34 = 246.80.
  10. Answer: The four decimal places claim precision the meter cannot deliver.
    The meter resolves pH only to 0.01 unit, and the two source readings were themselves given to one decimal place, so the blended value is uncertain well before the third decimal place. Reporting 6.3924 suggests the plant can distinguish a pH of 6.3924 from one of 6.3931, which no reading in this problem supports. The defensible report is 6.39, and the operating decision that follows from it is the same either way.

Differentiation

Support

Give students a worked example converting a pH of 7.0 to a concentration of 1.0 x 10^-7 moles per liter and back again, with each calculator keystroke listed, before they begin problem 1. Provide a place-value chart of powers of ten from 10^-9 to 10^-5 so the size of each answer can be checked against the chart rather than trusted blindly.

Extension

Ask students to redo problems 4 and 5 for a blend of one part well water to three parts brook water, then plot blended pH against the fraction of well water for five mixing ratios. The resulting curve, which bends rather than running straight, is a strong visual argument for why pH cannot be averaged, and it gives students a concrete reason to prefer working in concentration.

ELL supports

Pre-teach "concentration," "blend," "dose," and "soda ash" alongside a simple diagram showing two pipes joining into one. Provide a bilingual reference card giving the two relationships in words: pH is the negative logarithm of the concentration, and the concentration is ten raised to the negative pH. Allow problem 10 to be answered in two short sentences using the frame "The meter can only measure to ___, so reporting ___ is not supported."

Extensions

Have students find the pH range reported in their own utility's most recent consumer confidence report, convert the reported minimum and maximum into hydrogen ion concentrations, and describe how large the swing is as a factor rather than as a difference. Comparing that factor against the fifty-fold difference between the two fictional Ospringe sources gives students a sense of how tightly a real distribution system is usually held.