P=M

M

0

0

2P M

2 (M - P)

0

P > M

P = M

0

M

0

P < M

0

2P

M - 2P

P=0

0

0

M

EXAMPLE 6-1:

If P = 86 ppm as CaCO3, and if M = 118 ppm as CaCO3

Then, situation 2 (from Table 6-4) exists (P > M)

or P is greater than of M);

Hydroxyl = 2P - M = (2 x 86) - 118 = 54 ppm as CaCO3

Causticity = hydroxyl alkalinity as CaCO3 3

= 54 3 = 18 ppm as OH-

Carbonate = 2(M - P) = 2 X (118 - 86)

= 64 ppm as CaCO3

Bicarbonate = 0 ppm as CaCO3

Check: Total = 54 + 64 + 0 =118 ppm M alkalinity as CaCO3

Review of each situation in Table 6-4 provides this information, with situation:

1. The tests for P alkalinity and M alkalinity are equal. This means that all of

the alkalinity is due to hydroxyl ions. There is no carbonate or bicarbonate

present. (This is rare but occurs when a caustic solution is not exposed to

air.)

2. The P alkalinity is greater than one-half of the M alkalinity. This indicates

that there is hydroxyl and carbonate alkalinity, but no bicarbonate

alkalinity.

3. The P alkalinity is equal to one-half of the M alkalinity. This indicates that

all the alkalinity is due to carbonate. There is no bicarbonate alkalinity, and

the hydroxyl alkalinity is insignificant.

4. The P alkalinity is less than one-half of the M alkalinity. This indicates that

carbonates and bicarbonates are present.

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