Making the Grade ECI Exams: Math, Math, and

More Math Presented by:

Kent O. McIntosh Senior Industrial-waste Inspector L.A. County Sanitation Districts

CWEA P3S Training Riverside

Tuesday, August 9, 2016

Using the Calculator

Casio fx-115ES PLUS

SHIFT Cursor 10x

(-) S D DEL

Conversions

Conversions

Conversions

Casio fx-115ES PLUS

Cursor

Calculations

Making the Grade: Calculating a GPA

CLASS UNITS GRADE POINTS GRADE POINT

Reading 4.0 A 4 16 Writing 3.0 B 3 9

Arithmetic 2.0 C 2 4 Quantum

physics 3.0 F 0 0

TOTAL 12.0 29

∑ui ∑uigi

42.21229

≈==∑∑

i

ii

ugu

GPA C+

YOUR TURN Calculating a GPA

UNITS GRADE POINTS GRADE POINT

3 F 0 2 D+ 1.33 5 B- 2.61

TOTAL = TOTAL =

Calculating a GPA UNITS GRADE POINTS

GRADE POINT

3 F 0 0 2 D+ 1.33 2.66 5 B- 2.61 13.05

TOTAL = 10 TOTAL =

15.71

321

332211

uuugugugu

ugu

GPAi

ii

++++

==∑∑

57.110

71.15≈=

Pretreatment Facility Inspection: A Field Study Training Program

Appendix II: Pretreatment Arithmetic

Sample Problem 53

CWF (40 CFR 403.6)

CWF (40 CFR 403.6)

FT = average total daily flow FD = average daily dilution flow

CWF (40 CFR 403.6)

WASTESTREAM TYPE

FLOW (GPD)

DAILY MAX ZN LIMIT (MG/L)

Regulated 50,000 2.61 Regulated 20,000 1.33 Dilution 30,000 0

CWF (40 CFR 403.6)

++−++

+×+×

=000,30000,20000,50

000,30000,30000,20000,50000,20000,50

000,2033.1000,5061.2TC

Lmg /57.1000,100100,157

≈=

CWF

321

332211

uuugugugu

ugu

GPAi

ii

++++

==∑∑

57.110

71.15≈=

Lmg /57.1000,100100,157

≈=

Compare this answer,

to the GPA we calculated earlier,

Mass Loading (sample problem 17 from Grade-3 Study Guide)

The Given Solution

Mass Loading As a GPA

c1 = 25 mg/L c2 = 0 q1 = 40,000 GPD = 0.04 MGD qtotal = 45 MGD

Lmgqqc

ci

ii /...0222.045

004.025=

+×==

∑∑

But this is with no removal. If 45% is removed, then 55% (0.55) is left: 0.55 x 0.0222 = 0.0122 mg/L

YOUR TURN Mass Loading/Local Limit

(sample problem 16 from Grade-3 Study Guide)

YOUR TURN Mass Loading/Local Limit

(sample problem 16 from Grade-3 Study Guide)

c1 = 600 mg/L c2 = X q1 = 24 MGD q2 = 1.5 MGD c = 710 mg/L

21

2211

qqqcqcc

++

=

Mass Loading/Local Limit c1 = 600 mg/L c2 = X q1 = 24 MGD q2 = 1.5 MGD c = 710 mg/L

21

2211

qqqcqcc

++

=

5.1245.124600710

++•

=X

Using the Calculator

5.1245.124600710

++•

=X

SOLVE FOR X

Pretreatment Facility Inspection Appendix II: Pretreatment Arithmetic

Sample Problem 14

Pretreatment Facility Inspection Appendix II: Pretreatment Arithmetic

Sample Problem 14

THE GIVEN

SOLUTION

Inconsistent units

Maximum POTW BOD =

Lmg /5.5993034.8

000,150=

×

Mass Loading as a GPA Consistent units

POTW Capacity = 600 mg/L Industry Flow = 48,000 GPD = 0.048 MGD Industry BOD = 3500 mg/L POTW Flow = 30 MGD POTW BOD = 450 mg/L

Mass Loading as a GPA POTW Capacity = 600 mg/L Industry Flow = 48,000 GPD = 0.048 MGD Industry BOD = 3500 mg/L POTW Flow = 30 MGD POTW BOD = 450 mg/L

30048.030450048.03500

+×+×

=c > 600 ?

Mass Loading as a GPA POTW Capacity = 600 mg/L Industry Flow = 68,000 GPD = 0.068 MGD Industry BOD = 3500 mg/L POTW Flow = 30 MGD POTW BOD = 450 mg/L

30068.030450068.03500

+×+×

=c

c = 457 < 600 The industry will not cause problems.

Impact of Toxic Waste on Sewer System Waste-strength Monitoring

Pretreatment Facility Inspection Sample Problem 11

THE GIVEN SOLUTION

THE GIVEN SOLUTION

(CONTINUTED)

Impact of Toxic Waste on Sewer System Waste-strength Monitoring

As a GPA

c1 = 0.85 mg/L c2 = 10.5 mg/L q1 = 150,000 GPD q2 = 25,000 GPD Is the expected concentration (the “GPA”) equal to 3.2 mg/L?

Impact of Toxic Waste on Sewer System Waste-strength Monitoring

As a GPA

c1 = 0.85 mg/L c2 = 10.5 mg/L q1 = 150,000 GPD q2 = 25,000 GPD Is the expected concentration (c, the “GPA”) equal to 3.2 mg/L?

TRY THIS

Impact of Toxic Waste on Sewer System Waste-strength Monitoring

As a GPA

c1 = 0.85 mg/L c2 = 10.5 mg/L q1 = 150,000 GPD q2 = 25,000 GPD Is the expected concentration (c, the “GPA”) equal to 3.2 mg/L?

21

2211

qqqcqcc

++

=

2.2000,25000,150

000,255.10000,15085.0=

+•+•

=

Since the actual concentration (3.2 mg/L) is greater than the expected, there does appear to be a leak.

Concentration and Volume Grade-1 Sample Question 19

Concentration and Volume Grade-1 Sample Question 19

The Given Solution:

The final answer is wrong, and the equation is set up wrong. The correct equation and answer are:

25 𝑚𝑚 𝑥 0.01 𝑁100 𝑚𝑚 = 𝑁𝑓

𝑁𝑓 = 0.0025 𝑁

Concentration and Volume TRY THIS

What volume of a 3.2 N acid will neutralize 1.0 mL of a 7.9 N base?

Concentration and Volume

What volume of a 3.2 N acid will neutralize 1.0 mL of a 7.9 N base?

N1V1 = N2V2 3.2 x V1 = 7.9 x 1 V1 = 7.9/3.2 ≈ 2.5 mL

pH (hydrogen-ion concentration)

][1log]log[

][1log]log[

−−

++

=−=

=−=

OHOHpOH

HHpH

1410][10][

=+== −−−+

pOHpHOHH pOHpH

A change of 1 pH (or pOH) unit is a 10-fold change in concentration; for example, a pH-4 solution is 10 times more acidic than a pH-5 solution.

pH

THE GIVEN SOLUTION

What was your answer to this question?

What volume of a 3.2 N acid will neutralize 1.0 mL of a 7.9 N base?

What was the answer to this question?

pH problems are often simply concentration-volume problems

1.2)9.1114()14(

5.2

10101010][1010][

−−−−−−−

−−+

====

==pHOH

pH

OHH

q1 = ? q2 = 1 gpm

pH problems are often simply concentration-volume problems

1.2)9.1114()14(

5.2

10101010][1010][

−−−−−−−

−−+

====

==pHOH

pH

OHH

Since normality is a measure of concentration and since flow = volume/time, q = v/t

N1V1 = N2V2 ≡ N1V1/t = N2V2/t → c1q1 = c2q2

q1 = ? q2 = 1 gpm

pH problems are often simply concentration-volume problems

0079.010][0032.010][

1.22

5.21

===

===−−

−+

OHcHc

Since normality is a measure of concentration and since flow = volume/time, q = v/t

N1V1 = N2V2 ≡ N1V1/t = N2V2/t → c1q1 = c2q2

q1 = X q2 = 1 gpm

11010 1.25.2 •=• −− XSolve for X

Compare the difficulty of the solutions

11010 1.25.2 •=• −− X

Making the Grade with pH

baseacid

basebaseacidacidionexcess vv

vcvcc+

−+=±

)(

ba

bbaax vv

vcvcc+−

=±

If the term on the right side of the equation is negative, then the negative ion (OH-) is in excess, and the term on the left should also be negative.

Making the Grade with pH

ba

bbaax vv

vcvcc+−

=±

What if there is no excess ion; that is, the final mixture is neutral (pH 7)? Then cx = 0, and

ba

bbaa

vvvcvc

+−

=0

Which simplifies to

bbaa vcvc −=0

bbaa vcvc = which is just our earlier concentration-volume equation

Making the Grade with pH

ba

bbaax vv

vcvcc+−

=±

The Given Solution

Making the Grade with pH OUR SOLUTION

UNNECESSARY INFORMATION

Because the desired pH is acidic (pH < 7), the excess ion concentration is positive.

ba

bbaax vv

vcvcc+−

=+And if the pH of the caustic is 11.5, the pOH = 14 – 11.5 = 2.5

Making the Grade with pH OUR SOLUTION

UNNECESSARY INFORMATION

Because the desired pH is acidic (pH < 7), the excess ion concentration is positive.

ba

bbaax vv

vcvcc+−•

=+2 From the stoichiometric

equation, the acid has twice the concentration of hydrogen ions in a mole of the acid.

Making the Grade with pH OUR SOLUTION

UNNECESSARY INFORMATION

ba

bbaax vv

vcvcc+−•

=+2 cx = 10-6.5 mol/L

ca = ? = X va = 1000 gal cb = 10-2.5 mol/L vb = 45,000 gal TRY THIS

OUR SOLUTION Using the calculator

000,451000000,45101000210

5.25.6

+•−•

=−

− X

SOLVE FOR X

OUR SOLUTION

000,451000000,4510100010

5.25.6

+•−

=−

− X

X ≈ 0.071 M = ca

pH: Visualizing the Answer

A caustic tank (pH 10.5) is being emptied at 45 gpm. It’s combined with an acid-tank discharge (pH 0.5) so that the resulting combined discharge has a pH of 7.0. What is the acid-tank discharge rate?

THE GIVEN

SOLUTION

pH: Visualizing the Answer

How many pH units is the caustic (10.5) above neutral (pH 7)? How many units is the acid (pH 0.5) below neutral? What’s the difference between these two?

pH: Visualizing the Answer

How many pH units is the caustic (10.5) above neutral (pH 7)? 3.5 How many units is the acid (pH 0.5) below neutral? 6.5 What’s the difference between these two? 3

Knowing that a difference of one pH unit is a 10-fold difference in concentration, how much stronger is the acid than the caustic? 103 = 10 x 10 x 10 = 1000 So you would need only 1/1000 of the volume (or flow rate) of acid to neutralize the caustic. If the caustic is flowing at 45 gpm, what is the acid flow rate? 1/1000 x 45 = 0.045 gpm

THE GIVEN

SOLUTION

pH: Visualizing the Answer This technique can be

used when the pH of the two solutions differ by a whole number and when complete neutralization (to pH 7) occurs.

A modification of this technique can also be used even when complete neutralization does not occur.

pH: Visualizing the Answer

pH: Visualizing the Answer

THE GIVEN SOLUTION

pH: Visualizing the Answer

The acid is 5 pH units below neutral; the alkali is 4 units above. Because the acid is one unit farther from neutral than the alkali, it is 10 times more concentrated (101 = 10) and so only 1/10 as much is needed to neutralize the alkali: 1/10 x 500 = 50 gal. After neutralization, there will be 300 - 50 = 250 gal of pH-2 acid in a volume of 300 + 500 = 800 gal.

pH: Visualizing the Answer

The acid is 5 pH units below neutral; the alkali is 4 units above. Because the acid is one unit farther from neutral than the alkali, it is 10 times more concentrated (101 = 10) and so only 1/10 as much is needed to neutralize the alkali: 1/10 x 500 = 50 gal. After neutralization, there will be 300 - 50 = 250 gal of pH-2 acid in a volume of 300 + 500 = 800 gal. The concentration is simply:

Lmolc /003125.0800

25010 2

=×

=−

And the pH is –log(0.003125) = 2.5

MAKING THE GRADE

Combined Wastestream Formula Plant loading Surveillance monitoring pH (neutralization) problems The solutions to some simple pH problems can

be visualized without making any calculations An inexpensive, approved calculator can make

the math even easier

This “GPA” (weighted averages) method has been used for

MAKING THE GRADE

By making the grade (calculating weighted averages), you will get a high grade (at least on the math problems) when you take your appropriate Grade-level certification test.

GOOD LUCK!