pH AND ALKALINITY CONTROL

Introduction

The importance of obtaining correct pH measurements has greater emphasis today than at any other time. Its measurement is necessary both in the field and the laboratory as pH directly affects the functioning of drilling fluid additives.

pH

pH is the measurement of the relative degree of acidity or alkalinity of an aqueous solution. The numerical reading of the pH scale ranges from 1 (acid) to 14 (base), with 7 being the neutral point. It is known that the value of pH in the ranges stated are directly related to the effective or active acid concentration of a solution by the negative logarithm of the hydrogen ion concentration. The equation may be written as follows:

pH = - log [H+]

The [H+] in the equation represents the hydrogen ions in the solution, and can be referred to as the strength of the hydrogen ion in solution. If we have a strong acid of 0.01 molar, the pH is equal to 2 because the hydrogen ion concentration is as follows:

H+ is 10-2 molar pH = - log [10 -2] pH = 2

The pH scale is based on the dissociation constant of water. In distilled water, there are a few molecules that react with one another to form hydronium ions (H3 O+). This accounts for the acidic properties. The base properties are given by hydroxyl (OH -) ions of the solution. The following reaction is how they focus:

2H2O = H3O+ + OH- or H2O = H+ + OH-

Distilled water, at 25° C, will dissociate until the acid (H+) and the base (OH-) concentrations are equal. So the neutral pH value is as follows:

1 x 10-7 molar

The product of both concentrations is the equilibrium constant for water. It is referred to as the dissociation constant Kw:

KW = (H+) (OH-) Kw = [1 x 10-7] [1 x10-7] Kw = [1 x 10-14]

In this equation, the hydrogen (H+) ion concentration equals 1 x 10-7, thus the pH is 7. This is referred to as the neutral point.

In the continuous phase of drilling fluids, water is the primary aqueous solution. Drilling fluids allow us the flexibility of an increase in either the acid or base concentration, and

an increase in either concentration always results in a decrease in the other concentration.

To demonstrate how a strong base material will effect the hydroxyl ion concentration, a solution of strong caustic soda (0.01 NaOH) will be used in the example below:

H+ =Kw

[OH]

10 -14

= -2

10= 10 -12

pH = -Log [10-12] pH = 12

The pH chart in Figure No. 1 clearly shows the relationship of pH value to the activity of base/acid solutions. It must also be remembered that:

All concentrations are expressed in moles/liter (H+) x (OH-) = Kw = 1 x 10-14 moles/liter Kw = Equilibrium constant for water As (H+) decreases, (OH-) must increase if the product is to remain constant (1 x 10-14 moles/liter) = Kw

(H+) moles/liter pH (OH-) moles/liter

Very High 1 x 10-0 0 1 x 10-14 Very Low concentrati 1 x 10-1 1 1 x 10-13 Concentrati on ACID 1 x 10-2 2 1 x 10-12 on ACID

1 x 10-3 3 1 x 10-11

1 x 10-4 4 1 x 10-10

1 x 10-5 5 1 x 10-9

1 x 10-6 6 1 x 10-8

1 x 10-7 7 1 x 10-7

Very Low 1 x 10-8 8 1 x 10-6 Very High Concentrati 1 x 10-9 9 1 x 10-5 Concentrat on BASIC 1 x 10-10 10 1 x 10-4 ion BASIC

1 x 10-11 11 1 x 10-3

1 x 10-12 12 1 x 10-2

1 x 10-13 13 1 x 10-1

1 x 10-14 14 1 x 10-0

Figure 1

API recommends two methods for measuring the pH of drilling fluids. They are:

• Colorimetric method - paper strips impregnated with indicator dye. • accurate to only 0.5 pH unit • improper field storage, temperature and humidity affects dye • high salt concentrations affect readings (10,000 mg/L or more) • dark filtrates affect recognition

• Electrometric method - pH meter with a glass electrode • accurate to within ± 0.02 pH unit • can be standardized by calibrating with buffer solutions • temperature compensated • pH predicted by low voltage, which is very accurate • poor accuracy with saturated NaCl and KCl solutions

E =E o -

where:

2.303

F(RT) pH

E = measured voltageEo = total constant voltage in the measuring systemR = gas law constantT = temperature in OK G

F = Faraday's constant• Denotes for use in salt solutions (not saturated Cl- solutions)

pH is related to alkalinity, but it is not the same as alkalinity.

Alkalinity

Alkalinity may be defined as water soluble ions that neutralize acid. In other words, any ion that places a proton (H+) in the mud acts as an acid. On the other hand, any ion which combines with a proton (H+) is a proton acceptor and contributes to alkalinity. The following are some examples:

1. H SO 2 H+ 2-

+ SO The acid places 2 protons 2 4 4

Sulfuric Acid Hydrogen Sulfate (H+) in the water2. NaOH Na+ + OH- The hydorxide (OH-)

Sodium hydroxide Sodium Hydroxide combines with the protonOH- + H+ HOH (OH-)contributes to

Hydroxide Hydrogen Water alkalinity

There are basically three ions in water base muds that contribute to alkalinity:

Hydroxyl ions OH-

+ H+ HOH (water) Carbonate ions CO32- + H+ HCO3- (bicarbonate)Bicarbonate ions HCO + H+ CO + HOH (water)

3 2

In the above reactions, a proton (H+) combined with each ion. Therefore, each ion contributes to the alkalinity of a drilling fluid. Alkalinity and pH are related, but they are not the same.

Sources of Alkalinity

Caustic soda (NaOH), potassium hydroxide (KOH) and lime [Ca(OH)2 ] are added todrilling fluids at times to increase the pH of the fluid.

Cement is a common source of hydroxyl (OH-) ions. The chemical reaction is as follows:

Ca(OH) → Ca2+ 2OH-

Lime Calcium Hydroxyl

Anything added to the drilling fluid that places an OH- ion in solution will increase thepH.

The carbonate (CO3 2-) ion has several interesting sources in nature. It is sometimes foundin the form of potash (K CO ). It may also be the product of a chemical reaction taking

2 3

place as a result of a gas kick. When CO2 enters the drilling fluid, the following reactionoccurs:

1. CO2 + H2O→ H2CO3

Carbonic Acid

2. H CO + OH- → HCO -

+ H O 2 3 3 2

Bicarbonate

3. HCO + OH- → CO 2- + H O3 3 2

Carbonate

A third source of CO2 comes as a byproduct of bacterial action on certain types of organic matter. Fluids that contain mixtures of vegetable fibers (loss circulation materials) or polymers are subject to bacterial action. This is also true of fluids that contain certain types of organic filtration control additives. A prime example is corn or potato starch. A chemical biocide should be added in conjunction with the starch to control bacterial activity, or when the fluids contain biodegradable organic substances.

A fourth source of the carbonate/bicarbonate ion is make-up water. Any water used to mix or build drilling fluids should be checked for the presence of both ions prior to actual mixing of the fluid.

The bicarbonate (HCO -) ion generally occurs as the result of CO gas in the drilling3 2

fluid: The presence of the HCO3 - ion is dependent upon the pH or OH- concentration(See Figure 1.) Also, note the reaction below which occurs at pH of 10.3 and above.

HCO -3

Bicarbonate+ OH-

Hydroxyl

2-

→ HOH + CO3

Water Carbonate

The reaction shows that the bicarbonate ion cannot co-exist in a fluid of relative high pH. The hydroxyl ion reacts to produce carbonate and water. This concept is used to counteract bicarbonate contamination. The pH range in which the maximum amount of bicarbonate ions can be found is 7.0 - 10.5, as seen in Figure 2.

At a pH of 10.5 and above the OH- concentration becomes large enough that the above reaction predominates leaving CO3 2- and water. Occasionaly bicarbonate of soda can befound in the earth's formations, however the primary source is CO2 gas.

The following chart shows the predominate pH range for the alkalinity ions.

p H 4.3 - 6.3 Carbon dioxide (CO2 ) gas mainly in solutionpH 6.3 - 10.3 Bicarbonate (HCO3-) ion mainly in solutionpH 10.3> Carbonate (CO3 2-) ion mainly in solution

Knowledge of the relationship between pH and alkalinity can provide insight into whichion or ions are in solution and contributing to contamination problems. For example, it isvirtually impossibe to have CO or HCO - contamination if the pH is 11.5 - 12.0. On the

2 3

other hand, it is also highly unlikely to experience CO3 2- contamination if the pH is 8.5 -9.5.

Pf Alkalinity

The Pf titration is performed on the mud filtrate. Phenolphthalein is used as the indicatorand N/50 (0.02N) sulfuric acid (H SO ) is the titrating solution. The endpoint for the

2 4

titration occurs when the color of the sample being titrated turns from reddish-pink backto the original color of the filtrate. The color change occurs at a pH of 8.3, and Pf isreported as the number of milliliters of N/50 sulfuric acid used to reach the endpoint.

Noting the pH ranges listed above, it is readily seen that in the Pf titration the two ionsthat predominate are CO 2- and OH-, with a small quantity of HCO - ions present.

3 3

Therefore the two reactions which occur in the Pf titration are:

1. CO 2- + H SO (N/50)3 2 4

Carbonate Acid → HCO - + HSO -3 4

Bicarbonate Bisulfate2. OH- + H SO (N/50) → HOH + SO 2-

2 4

Hydroxyl Acid4

Water Sulfate

K d J

�= +

A

♦ Ww V

W

O U

1LJU, A

116

0 N 4

C o Ov

xC.

N+

48

W

Y

Q U .H

LU

-d

O V C

Figure I

CA 14P■ ■

(j

IL IC6W W W +Y

$-Z

W J

d?

C Oo 40

Figure 2

0

In other words, at the 8.3 pH endpoint the following is true:

Most of the CO 2- has been converted to HCO - as seen in Reaction No. 1, and most of3 3

the OH- has been converted to HOH as seen in Reaction No. 2. Therefore, it is possibleto conclude that when there is a high concentration of either OH- or CO3 2- ions, the Pfwill be rather large.

The question is then how does one tell which ion is present? In order to determine whichof the two ions is present in the P titration, the M titration must be conducted.

f f

Of the two ions (OH- and CO 2-), only CO 2- will affect the M3 3 f

titration. This is readily

seen in Reaction No. 1, as the product produced is HCO3 - The bicarbonate ion will thenmanifest itself in the Mf titration.

The following reaction occurs when conducting the second titration to determine Mf:

3. HCO -

+ H SO → CO + H O + HSO4-3 2 4 2 2

Bicarbonate Acid Carbon Water BisulfateDioxide

In Reaction No. 3, most of the HCO - is converted to CO at the 4.3 pH endpoint. The3 2

HCO3 - ion determined in this titration may have come from the titration in Reaction No.1 or the HCO3 - ion may have been present in the filtrate prior to conducting any of thetitrations. In other words, high concentrations of CO 2- ions in the P titration will

3 f

produce high concentrations of HCO - ions in the M titration as seen below:

P titration converts CO3 f

-2- to HCO

f 3 3

M titration converts HCO - to COf 3 2

Mf Alkalinity

After performing the the Pf titration, add methyl orange or bromo-cresol green/methyl orange and continue to titrate with N/50 sulfuric acid. If methyl orange is used as the indicator, the sample will change from orange to salmon-pink when the endpoint is reached. If bromo-cresol green is used as the indicator, the sample will change from blue-green to yellow. The pH of the sample should be approximately 4.3 when the endpoint is reached. This is the second titration, and the milliliters of acid used in the titration are added to the milliliters of acid required to reach the Pf endpoint and the totalamount of acid used is recorded as Mf alkalinity. The following is an example titration ofa filtrate with a pH of 10.4.

pH = 10.4

Pf = 3.2 ml N/50 H2SO4 pH = 8.3 Color change: pinkto original color

M = 5.3 ml N/50 H SO4f 2

(3.2 ml + 2.1 ml to reduce pH from 8.3 to 4.3) pH = 4.3 Color changes from

orange to pink orblue to yellow

In the example above, a total of 5.3 ml of N/50 sulfuric acid was required to reduce thepH from 10.4 to 4.3. The P alkalinity is reported as 2.1 and the M alkalinity is reported

f

as 5.3 (3.2 + 2.1 = 5.3).

Using the following diagram it is possible to examine P

f

and M in terms of the ionsf

present in a given titration:

pH OH- ions, OH- + CO 2- ions, or CO 2- ions3 3

11.48.3 OH- ions, OH- + CO 2- ions, or CO 2- ions + HCO

f

Filtrate pH

- ions P 3 3 3 f

Filtrate pH4.3 HCO - ions and CO2 gas M

3 f

Interferring ions Filtrate pH

Knowing that CO 2- effects both the Pf and the M, it is easy to distinguish which ion is3

present in the Pf.

OH- will not be present in the M f

f

titration

High concentrations of OH- and HCO3

following reaction:

2-

- cannot exist simultaneously. This is due to the

HCO - + OH- → HOH + CO3 3

If the P and M values are both high, the main contaminant has to be CO 2- If the P isf f 3 f

high and the M is relatively unchanged, the predominate ions are OH- ions. If the Pf

value is relatively small and the Mf

value is large, the predominate ions are HCO - ions. f 3

Relating these values to pH can be easy. When the pH is greater than 10.5 and the Pf andM values are relatively high, the predominating ions will be CO 2- ions. If the pH is less

f 3

than 10.0 and the M value is large, the predominate ions will be HCO - ions. If the pH isf

greater than 10.0, P is high and M3

is virtually unchanged, the predominate ions are f f

most likely OH- ions. HCO - and CO 2- ions can co-exist in certain pH ranges, therefore,3 3

knowledge of the pH can aid in determining which of the two ions is causing the problem.

Estimating the Concentration of OH-, HCO3- and CO32- Ions in Water Base Drilling Fluids

If a red color does not develop upon the addition of phenolphthalein to a sample offiltrate, the P is recorded as zero (0). (See Figure No. 3.) If P = 0, the pH is <8.3, and

f f

the ions contributing to alkalinity are HCO - ions which will be revealed by the M3 f

titration and reported as follows:

OH- = 0CO32- = 0HCO -, meq/L = M x 20

3 f

If a pink or yellow color does not develop when methyl orange or bromo-cresol green isadded to a sample of filtrate following the titration for Pf, additional titration with acid isnot required as the pH is already at 4.3. When this occurs the ions contributing toalkalinity are OH- ions only. (If CO 2- ions were present, titration for M would have been

3 f

required, and its not.) The ions are reported as follows:

OH-, meq/L = Pf x 20CO32- = 0HCO3- = 0(When a second titration is not required, P = M.)

f f

The following expressions are used to determine the ions contributing to alkalinities.

P = 0 The alkalinity is all HCO - ionsf 3

P = M The alkalinity is all OH- ionsf f

2 x Pf> M The alkalinity is a mixture of CO 2- and OH- ionsf 3

2 x P = M The alkalinity is all CO 2- ionsf f 3

2 x P < M The alkalinity is a mixture of HCO - and CO 2- ionsf f 3 3

The following explains the reasoning behind the statements above:

Pf = O

⎧ no (OH-)P

f= O ⎨

⎩ no (CO3

⎧ -

2- )

8.3 pH

2nd =3 ml ⎨⎩

HCO 3

4.3 pH

Upon adding the phenolphthalein, there was no color change. This means the pH is already at 8.3 or less. Therefore, there cannot be any OH- or CO3 2- ions in the filtrate.The only ions that can be in the second titration will be HCO3 -.

0 ml = Pf

+ 3ml = 2nd titration3 ml = M f

(OH-) me/L = 0(CO3 2-) me/L = 0(HCO -) me/L = 20 x M

3 f

P = Mf f

P f = 2 ml

2nd =0

⎧ (OH-) only ⎨

⎩ no (CO32-) 8.3 pH-

⎧ no (HCO3 ) ⎨

⎩ 4.3 pH

Upon adding phenolphthalein, a red color develops indicating possible OH- or CO32-

ions. After titrating to the 8.3 pH endpoint, methyl orange is added and immediately asalmon pink color results. This simply means that the 8.3 pH endpoint is the same as the4.3 pH endpoint. In other words, there is no second titration. Therefore, there can be noHCO - ions. Also, there can not be any CO 2- ions. CO 2- ions would have been converted

3

to HCO - in the P titration, and a M3 3

titration would have followed. 3 f f

(OH-) me/L = Pf x 20(CO3 2-) me/L = 0(HCO

3 -) me/L =

2 x Pf > Mf

⎧Pf = 2 ml ⎨

⎩

0

OH -

2

CO3

→ H 2O-

-

→ HCO3

8.3 pH

HCO -

→ CO 3 2

Since most of the acid was used in the P, it is likely that OH- and CO 2- were present.f 3

Since CO 2- shows up in the 2nd titration, it is likely there is some CO 2- because there3 3

was a small titration. With HCO3 - present, the 2nd titration would have been muchlarger.

2nd = 1 ml

4.3 pH Pf = 2 ml2nd = 1 mlM

f

(OH-) me/L = 3 ml = (40 x P) - (20 x M)

f f

(CO 2-) me/L = 40 (M - P)3 f f

HCO3 - me/L = 02 x P > M

f f

Pf =1 ml

⎧

⎪

⎪

⎨

CO 3

2-

-

→ HCO 3-

⎪ HCO ⎪⎩

⎧ HCO

3

-

→ CO2

from Pf

→ CO2nd =3 ml ⎨

⎩3 2

(present before Pf )

Since the second titration was large, it is sure that HCO - is present. The P has to be3 f

mostly CO 2- because the OH- would have reacted with the HCO - to form CO 2- + H O.3

The HCO - produced in the P3 3 2

shows up in the M. 3 f f

Pf

2nd

M f

= 1 ml= 3 ml = 4 ml

Remember: HCO - that shows up in the Mtitration can come from two sources:

3 f

The titration of the CO 2- ion in the P yields HCO - ion.3 f 3

There may be some HCO - ions in the filtrate prior to running the P or M test.3 f f

(OH-) = 0(CO 2-) me/L = 40 x P

3 f

(HCO -) = (20 x M) - (40 x P)3 f f

2 x Pf = Mf

P = 2 ml {CO 2- -→ HCO

f 3

2nd titration

{HCO

= 2 ml

-3

3

→ CO2

In the case where 2 x P= M,it is easy to see that the only ion present is CO 2- If the

f f

alkalinity was solely from HCO - then the P3

would have been small because HCO - does 3 f 3

not effect the Pf. The Pf could not have been from OH- and HCO3 -, as they cannot co-exist in high concentrations at the same time.

2 Mf = Pf

+ 2 ml = 2nd4 ml = M f

In other words, all CO 2- in the Pwas converted to HCO - in the P titration. The HCO -

3

was then converted to COf 3 f 3

in the M titration. 2 f

(OH-) = 0(CO

3 2-) = 40 x pf (HCO3 -) me/L = 0

The alkalinity titrations discussed here are normally recorded on the daily drilling muil report. It is necessary to keep these values in line for the type system that is run, or a great number of problems may occur.

P1 P2 Alkalinity

In determining P1 , 2 ml of N/10 (0.1 N) NaOH is added to the mud filtrate. The NaOH isrequired to raise the pH above 11.4. Above 11.4 pH, any HCO3 - will be converted toCO3 2- via the following reaction:

HCO -

+ OH- → CO 2- + HOH 3 3

Bicarbonate NaOH Carbonate Water2 mls N/10

The next step is to add Barium Chloride (BaCl2 ). Remember, never pipette bariumchloride as it is toxic. The barium (Ba) reacts with CO 2- to form insoluble BaCO

3 3

(barium carbonate):

CO2-

+ BaCl → BaCO ↓ + Cl-3 2 3

Carbonate Barium BariumChloride

Chloride carbonate

This leaves only OH- in solution. The concentration of OH- will equal the original concentration of OH- in the filtrate plus that added via N/10 NaOH. From the total (OH-) concentration, one must now subtract the OH- ions used in the conversion of HCO3 - toCO3 2-. Emperically it looks like this:

Hydroxyl Concentration(OH-) + (OH-)Original From

In N/10Filtrate NaOH

added

→ (OH-) + (OH-)Used in Afterconversion addingof BaCl 2

2-

HCO - to CO

The OH- concentration remaining after the addition of BaCl 2

with (N/50) hydrochloric acid.

3 3

in step 2, is then titrated

OH- + HCl → HOH + Cl-

Hydroxyl N/50 Water ChlorideAfter BaCl2 hydrochloricis added acid

In determining P , the exact same procedure for determining P is followed, except that2

the mud filtrate is left out. P1

measures the total OH- which was added in P .2

If P > P , this indicates that none of the OH- ions in P

1

were consumed in the conversion 1 2 1

of HCO - to CO 2-. Therefore the only ions in the filtrate are CO 2- + OH-. Since the CO 2-3 3 3 3

was removed by the BaC12 , the OH- concentration is equal to the difference in theamount of acid used in P and in P .

1

So, me/L (OH-) = 20 x (P

2

- P ) or mg/L(OH) = 340 (P - P ) 1 2 1 2

Since the P gives a total amount of HCl to neutralize both the CO 2- and the OH-, thenf

ME/L CO 2- = P - (P - P ) x 40 or mb/L CO

3

2- = 1200 [Pf - (P - P )} 3 f 1 2 3 1 2

The me/L of acid used in neutralizing the CO3 2- must be multiplied by 2 since thecarbonate ion has a -2 charge. Therefore, the multiplying factor of Mg/L for CO32-

becomes 40 instead of 20. To covert to mg/L of CO3 2- multiply the me/L x 60/2 (Eqwt ofCO3 2-.

If P1 < P , this indicates that some of the N/10 NaOH added in Pwas consumed in the

2

following reaction:

NaHCO + NaOH → Na

1

CO + HOH 3 2 3

Sodium + Sodium Sodium WaterBicarbonate hydroxide carbonate

Since P = total OH- concentration added in P , the P - P = the amount of acid (ml) used 2 1 2 1

in converting the HCO - to CO 2-. Therefore,3 3

me/L HCO-= 20 x (P - P) or mg/L HCO 2- = 1200 P

3 2 l

It has been established that 40 x P3

- (P - P ) = me/L COf

2-. However, one cannot ave f f 2 3

HCO - and OH- ions simultaneously in the mud filtrate, only HCO - and CO 2-.3

If OH- = 0, then P - P3 3

= 0, and the equation for determining the me/L CO 2- becomes;1 2

CO 2- = 40 x P - (0), since P - P3

= 0. This simplifies to 40 x P.3 f

So, me/L CO 2- = 40 x P

1 2 f

when P < P . Or mg/L CaCO 2- = 1200 P 3 f 1 2 3 f

The P - P is a alternate test but it has a standard place on the daily drilling mud report.1 2

It is not normally run unless an alkalinity problem may exist. (BARIUM CHLORIDE TITRATING SOLUTION, IF INGESTED, CAN KILL YOU).