Determine the ionic strength in solution based on TDS
I = (2.5 × 10 −5 ) × TDS(mg/L)
I = 0.00448
Determine activity coefficient for HCO3-
0.5(z i )2 I 1/2
logγi = −
1+ I 1/2
logγHCO −
3
0.5( −1)2 (0.00448)1/2
=−
= −0.0313
1+ (0.00448)1/2
γHCO − = 0.93
3
Determine activity coefficient for Ca+2
logγCa +2 = −
0.5( +2)2 (0.00448)1/2
= −0.125
1+ (0.00448)1/2
γCa +2 = 0.75
SOLUBILITY PRODUCT
Calculate the solubility product of anhydrite at 25°C
CaSO4(s) ↔ Ca2+ + SO42Look up:
http://inside.mines.edu/~epoeter/_GW\19WaterChem3/GibbsFreeEnergyValues.xls
Δf G°Ca2+ = -553.6 kJ mol-1
Δf G°SO42- = -744.0 kJ mol-1
Δf G°CaSO4(s) = -1321.8 kJ mol-1
R (gas constant) = 8.314 J K-1 mol-1
so:
ΔrG° = -553.6 + ((-744.0)) - ((-1321.8)) = 24.2 kJ mol-1
logK SP =
− ΔrGo
− 24,200 J mol−1
=
= −4.24
2.303RT 2.303 (8.314 J K −1 mol −1 ) (298.15 K)
K SP = 10 − 4.24 = 5.75 x10 −5
assume Kelvin for thermodynamic expressions, unless noted otherwise
1
Suppose a groundwater analysis indicates
5x10-2 mol/L Ca2+ and 7x10-3 mol/L SO42- (assume TDS = 300 mg/L)
Is this water saturated with respect to anhydrite?
Determine the ionic strength in solution based on TDS
I = (2.5 × 10 −5 ) × TDS(mg/L)
I = 0.0075
Determine activity coefficient for both Ca+2 and SO4-2
logγCa +2 = logγSO −2 = −
4
0.5((+2)) 2 ((0.0075))1/2
= −0.16
0 16
1 + (0.0075)1/2
γCa +2 = γSO −2 = 0.69
4
Suppose a groundwater is analysis indicates
5x10-2 mol/L Ca2+ and 7x10-3 mol/L SO42- (assume TDS = 300 mg/L)
Is this water saturated with respect to anhydrite?
CaSO4(s) ↔ Ca2+ + SO42-
(
) (
⎛ IAP ⎞
⎛ Ca 2+ act SO 24−
⎟⎟ = log
SI = log
l ⎜⎜
l ⎜⎜
K SP
⎝ K SP ⎠
⎝
[
] [
]
)
act
⎞
⎟⎟
⎠
⎛ γ Ca Ca 2+ γ SO4 SO 24− ⎞
⎟
= log⎜
⎜
⎟
K
SP
⎝
⎠
-2
-3
-4
IAP = 0.69(5x10 )0.69(7x10 ) = 1.68x10 = 10-3.77 mol2 L-2
4 24 moll2 L-2
2
KSP = 10-4.24
⎛ IAP ⎞
⎛ 10 −3.77 ⎞
⎜
⎟
SI = log⎜
⎟ = log⎜⎜ 10 − 4.24 ⎟⎟ = 0.47
K
⎝
⎠
⎝ SP ⎠
In this case, SI > 0, i.e., IAP > KSP, so the solution is
supersaturated and anhydrite should precipitate
2
Consider the acid-base reaction
H2CO3 ↔ HCO3- + H+
K = 10-6.35
Which way should the reaction go if
a H2CO3 = 10-4
pH = 7
a HCO3- = 10-3?
First, calculate the IAP
Recall
pH = -log a H+
IAP =
aHCO − aH+
3
aH CO
2
3
a H+ = 10-7
so
=
(10 )(10 ) = 10
(10 )
−3
−7
−6
−4
IAP > K (10-6 > 10-6.35)
The reaction will shift to the left until IAP = K
so more H2CO3 will be formed
ALKALINITY
A 100 mL sample with a pH of 8 is titrated to the methylorange-end-point (which goes from yellow at pH 4.4 to red at
pH 3.1) with 2 mL of 0.5 N H2SO4
Whatt iis th
Wh
the total
t t l alkalinity
lk li it in
i mg L-11 as CaCO
C CO3 and
d what
h t is
i the
th
-1
concentration of HCO3 in mg L ?
The total alkalinity in mg L-1 as CaCO3 is given by:
2 mL × 0.5 eq/L × 50 g/eq × (1000 mg/g)
100 mL
= 500 mg/L as CaCO3
Alk T =
The concentration of HCO3- is given as:
2 mL × 0.5 eq/L × 61 g/eq × (1000 mg/g)
100 mL
= 610 mg/L
HCO3- (mg/L) =
3
CORRECTION on Check Correctness of Analysis:
Calculate TDS
Calculated TDS =
0.6Alkalinity + Na + K + Ca + Mg +
Cl + SO4 + SiO2 + NO3-N + F
Solute
Measured
Conc (mg/L)
Ca2+
92.0
M 2+
Mg
34 0
34.0
0 6Alk li it because
0.6Alkalinity
b
CO3 60% off CaCO
C CO3 by
b weight
i ht
Na +
8.2
K+
1.4
Fe(III)
0.1
Reported Alkalinity OR
[HCO3-] * 50g/eq CaCO3 = 266.4mg/L
61g/eq HCO3 -
HCO3 -
325.0
SO4
2-
84.0
Cl -
9.6
NO3 -
13.0
Calculated
What
is theTDS
Calculated
= 437.2TDS?
mg/L
Measured TDS > Calculated TDS because some
species not included in the calculation (e.g. Fe+3)
acceptable range : 1.0 <
If the ratio is out of the
range, reanalyze
measured TDS
< 1.2
calcuated TDS
ratio =value
1.38 in
is this
not acceptable
Measured
case 603.5 mg/L
What is the ratio? Is it acceptable?
4