Find: Mg S O4 mg (hypothetical) L Ca2+ SO4 Mg2+ Na+ - HCO3 Cl- Ion C

Find: Mg S O4 mg (hypothetical) L Ca2+ SO4 Mg2+ Na+ - HCO3 Cl- Ion C
90 120 147 340 Ca2+ 98 Ca2+ SO4 2- Mg2+ 22 Mg2+ Na+ 83 Na+ HCO3 - - HCO3 317 Find the concentration of magnesium sulfate, in milligrams per liter. [pause] In this problem, --- Cl- 2- SO4 125 Cl- 89

90 120 147 340 Ca2+ 98 Ca2+ SO4 2- Mg2+ 22 Mg2+ Na+ 83 Na+ HCO3 - - HCO3 317 An aqueous solution with known concentrations of various ions, is provided. Cl- 2- SO4 125 Cl- 89

90 120 147 340 Ca2+ 98 Ca2+ SO4 2- Mg2+ 22 Mg2+ Na+ 83 Na+ HCO3 - - HCO3 317 These ions would hypothetically join, and form compounds, based on the equivalent concentrations, of each ion. Cl- 2- SO4 125 Cl- 89

90 120 147 340 Ca2+ 98 Ca2+ SO4 2- Mg2+ 22 Mg2+ Na+ 83 Na+ - - HCO3 HCO3 317 And magnesium sulfate is the compounds we want to calculate. [pause] We’ll solve this problem by creating a --- Cl- 2- SO4 125 Cl- 89

Find: Mg S O4 mg (hypothetical) L - Ion C Ca2+ 98 SO4 Ca2+ Mg2+ Na+
HCO3 2- Cl- Mg2+ 22 - Na+ 83 - HCO3 317 bar graph, based on the equivalent concentration of each cation, against the equivalent concentrations of each anion, from the solution. 2- meq L SO4 125 Cl- 89

HCO3 2- Cl- Mg2+ 22 - Na+ 83 - HCO3 317 The amount of magnesium sulfate which forms will depend on the amount of overlap, or shared concentration, between the magnesium ion and sulfate ion, as shown on the plot. 2- meq L SO4 125 Cl- 89

HCO3 2- Cl- Mg2+ 22 - Na+ 83 - HCO3 317 Therefore, we’ll first need to determine the equivalent concentrations of each ion, --- 2- meq L SO4 125 Cl- 89

HCO3 2- Cl- Mg2+ 22 - Na+ 83 - HCO3 317 and we’ll use the units of milliequivalent, per liter. [pause] Starting with the given concentrations --- 2- meq L SO4 125 Cl- 89

HCO3 2- Cl- Mg2+ 22 - Na+ 83 - HCO3 317 we’ll expand the table, to include columns for --- 2- meq L SO4 125 Cl- 89

Find: Mg S O4 mg (hypothetical) L Ion C Z C Ca2+ 98 Mg2+ 22 Na+ 83 -
ionic mg*Z meq L Ion C Z C weight meq Ca2+ 98 Mg2+ 22 Na+ 83 the ionic weight of each ion, the absolute value --- - HCO3 317 2- SO4 125 Cl- 89

ionic mg*Z meq L Ion C Z C weight meq Ca2+ 98 Mg2+ 22 Na+ 83 of the valence, z, and the calculated concentrations, --- - HCO3 317 2- SO4 125 Cl- 89

ionic mg*Z meq Ion C Z C weight meq L Ca2+ 98 Mg2+ 22 Na+ 83 in milliequivalent per liter. [pause] For clarity, we’ll begin by identifying --- - HCO3 317 2- SO4 125 Cl- 89

Find: Mg S O4 mg (hypothetical) L 98 22 83 317 125 89 Ion Ca2+ Mg2+
ionic Ca2+ Mg2+ Na+ HCO3 - SO4 2- Cl- weight C mg*Z meq Z cations the cations, from the anions. Then, the ionic weights are looked up anions

Find: Mg S O4 mg (hypothetical) L Ion C Z C Ca2+ 98 40.1 cations Mg2+
ionic mg*Z meq meq L Ion C Z C weight Ca2+ 98 40.1 cations Mg2+ 22 24.3 Na+ 83 23.0 from a table of values, and inserted here. [pause] Next, the value of Z for each ion equals --- - HCO3 317 61.0 2- anions SO4 125 96.1 Cl- 89 35.5

Find: Mg S O4 mg (hypothetical) L Ion C Z C Ca2+ 98 40.1 cations Mg2+
ionic mg*Z meq meq L Ion C Z C weight Ca2+ 98 40.1 cations Mg2+ 22 24.3 Z=|Valence| Na+ 83 23.0 the absolute value of the valence, for each ion. So calcium, plus 2, --- - HCO3 317 61.0 2- anions SO4 125 96.1 Cl- 89 35.5

Find: Mg S O4 mg (hypothetical) L Ion C Z C Ca2+ 98 40.1 2 cations
ionic mg*Z meq meq L Ion C Z C weight Ca2+ 98 40.1 2 cations Mg2+ 22 24.3 Z=|Valence| Na+ 83 23.0 has as Z value of 2, and Sodium, plus 1, ---- - HCO3 317 61.0 2- anions SO4 125 96.1 Cl- 89 35.5

ionic mg*Z meq meq L Ion C Z C weight Ca2+ 98 40.1 2 cations Mg2+ 22 24.3 Z=|Valence| Na+ 83 23.0 1 has as Z value of 1. The remaining Z values are entered. - HCO3 317 61.0 2- anions SO4 125 96.1 Cl- 89 35.5

ionic mg*Z meq meq L Ion C Z C weight Ca2+ 98 40.1 2 cations Mg2+ 22 24.3 2 Na+ 83 23.0 1 The milliequivalent per liter is calculated by multiplying --- - HCO3 317 61.0 1 2- anions SO4 125 96.1 2 Cl- 89 35.5 1

ionic mg*Z meq meq L Ion C Z C weight Ca2+ 98 40.1 2 cations Mg2+ 22 24.3 2 mg Na+ 83 23.0 1 C Z * meq L L The concentration, in milligrams per liter, by the Z variable, divided by the ionic weight. So for the calcium ion, this would be --- - HCO3 317 C 61.0 = 1 ionic mg*Z 2- anions SO4 125 96.1 2 weight meq Cl- 89 35.5 1

ionic mg*Z meq meq L Ion C Z C weight Ca2+ 98 40.1 2 cations Mg2+ 22 24.3 2 mg Na+ 83 23.0 1 98 2 * - meq L L 98 milligrams per liter times 2, divided by 40.1, milligrams per milliequivalent, the Z’s and milligrams cancel, and the concentration equals --- HCO3 317 C 61.0 = 1 mg*Z 2- anions 40.1 SO4 125 96.1 2 meq Cl- 89 35.5 1

Find: Mg S O4 mg (hypothetical) L Ion C Z C Ca2+ 98 40.1 2 4.89
ionic mg*Z meq meq L Ion C Z C weight Ca2+ 98 40.1 2 4.89 cations Mg2+ 22 24.3 2 mg Na+ 83 23.0 1 98 2 * - meq L L 4.89 milliequivalents per liter. [pause] The other ions are are calculated the same way. HCO3 317 C 61.0 = 1 mg*Z 2- anions 40.1 SO4 125 96.1 2 meq Cl- 89 35.5 1

ionic mg*Z meq meq L Ion C Z C weight Ca2+ 98 40.1 2 4.89 cations Mg2+ 22 24.3 2 1.81 Na+ 83 23.0 1 3.61 [pause] If we remove the middle 3 columns of the table, --- - HCO3 317 61.0 1 5.20 2- anions SO4 125 96.1 2 2.60 Cl- 89 35.5 1 2.51

ionic mg*Z meq meq L Ion C Z C weight Ca2+ 98 40.1 2 4.89 cations Mg2+ 22 24.3 2 1.81 Na+ 83 23.0 1 3.61 we’ll be are ready to construct the --- - HCO3 317 61.0 1 5.20 2- anions SO4 125 96.1 2 2.60 Cl- 89 35.5 1 2.51

Find: Mg S O4 mg (hypothetical) L Ion C Ca2+ 4.89 cations Mg2+ 1.81
meq L Ion C Ca2+ 4.89 cations Mg2+ 1.81 Na+ 3.61 milliequivalent per liter bar graph. - HCO3 5.20 2- anions SO4 2.60 Cl- 2.51

meq L Ion C Ca2+ 4.89 cations Mg2+ 1.81 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- Na+ 3.61 With the cations, on top, in green, --- - HCO3 5.20 2- anions SO4 2.60 Cl- 2.51

Find: Mg S O4 mg (hypothetical) L Ion C Ca2+ 4.89 cations cations Mg2+
meq L Ion C Ca2+ 4.89 cations cations Mg2+ 1.81 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- Na+ 3.61 and the anion, on bottom, in orange. The length of each ion represented on the graph, is based on --- - HCO3 5.20 2- anions SO4 2.60 anions Cl- 2.51

meq L Ion C Ca2+ 4.89 4.89 1.81 3.61 cations Mg2+ 1.81 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- Na+ 3.61 its concentration in milliequivalents per liter. If the left end of the graph is zero, then --- - HCO3 5.20 2- anions SO4 2.60 5.20 2.60 2.51 Cl- 2.51

meq L Ion C meq L Ca2+ 4.89 4.89 6.70 10.31 cations Mg2+ 1.81 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- Na+ 3.61 the milliequivalents per liter, between each ion, can be calculated. Since the question asks to find the hypothetical --- - HCO3 5.20 2- anions SO4 2.60 5.20 5.80 10.31 Cl- 2.51 meq L

meq L Ion C meq L Ca2+ 4.89 4.89 6.70 10.31 cations Mg2+ 1.81 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- Na+ 3.61 formation of magnesium sulfate, we’ll focus on the overlap between the ions, --- - HCO3 5.20 2- anions SO4 2.60 5.20 5.80 10.31 Cl- 2.51 meq L

meq L Ion C meq L Ca2+ 4.89 4.89 6.70 10.31 cations Mg2+ 1.81 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- Na+ 3.61 magnesium plus 2, and the sulfate minus 2. - HCO3 5.20 2- anions SO4 2.60 5.20 5.80 10.31 Cl- 2.51 meq L

Find: Mg S O4 mg (hypothetical) L MgSO4=6.70-5.20 Ion C Ca2+ 4.89
meq L Ion C meq 5.20 6.70 L Ca2+ 4.89 4.89 10.31 cations Mg2+ 1.81 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- Na+ 3.61 6.70, milliquivalents per liter, minus, 5.20, milliequivalents per liter, equals, --- - HCO3 5.20 2- anions SO4 2.60 5.20 5.80 10.31 Cl- 2.51 meq L

Find: Mg S O4 mg (hypothetical) L MgSO4=6.70-5.20 Ion C MgSO4=1.50
meq L meq Ion C MgSO4=1.50 meq L 5.20 6.70 L Ca2+ 4.89 4.89 10.31 cations Mg2+ 1.81 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- Na+ 3.61 1.50 milliequivalents per liter. [pause] Since the problem asks for the concentration milligrams per liter, --- - HCO3 5.20 2- anions SO4 2.60 5.20 5.80 10.31 Cl- 2.51 meq L

meq L meq L mg meq Ion C MgSO4=1.50 * 60.2 Ca2+ 4.89 cations Mg2+ 1.81 4.89 6.70 10.31 Na+ 3.61 we’ll multiply by the equivalent weight of magnesium sulfate, 60.2, milligrams per milliequivalent, and the hyptothetical concentration of --- - HCO3 5.20 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- 2- anions SO4 2.60 Cl- 2.51 5.20 5.80 10.31

meq L meq mg Ion C MgSO4=1.50 * 60.2 meq L Ca2+ 4.89 mg MgSO4=90.3 L cations Mg2+ 1.81 4.89 6.70 10.31 Na+ 3.61 magnesium sulfate equals 90.3 milligrams per liter. - HCO3 5.20 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- 2- anions SO4 2.60 Cl- 2.51 5.20 5.80 10.31

meq L meq mg Ion C MgSO4=1.50 * 60.2 meq L Ca2+ 4.89 mg 90 120 147 340 MgSO4=90.3 L cations Mg2+ 1.81 4.89 6.70 10.31 Na+ 3.61 When reviewing the possible solutions, --- - HCO3 5.20 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- 2- anions SO4 2.60 Cl- 2.51 5.20 5.80 10.31

Find: Mg S O4 mg (hypothetical) L AnswerA MgSO4=6.70-5.20 Ion C
meq L meq mg Ion C MgSO4=1.50 * 60.2 meq L Ca2+ 4.89 mg 90 120 147 340 MgSO4=90.3 L cations Mg2+ 1.81 AnswerA 10.31 Na+ 3.61 the answer is A. [pause] If the problem asked to find --- - HCO3 5.20 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- 2- anions SO4 2.60 Cl- 2.51 5.20 5.80 10.31

Find: Mg S O4 mg (hypothetical) L Hardenss? Ion C Ca2+ 4.89 cations
meq L Ion C Ca2+ 4.89 cations Mg2+ 1.81 4.89 6.70 10.31 Na+ 3.61 the hardness of the solution, as calcium carbonate, then the method would be to calculate the combined equivalent concentration of all multivalent cations, --- - HCO3 5.20 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- 2- anions SO4 2.60 Cl- 2.51 5.20 5.80 10.31

Find: Mg S O4 mg (hypothetical) L Hardenss? Ion C Ca2+ 4.89 cations
meq L Ion C Ca2+ 4.89 cations Mg2+ 1.81 4.89 6.70 10.31 Na+ 3.61 which for this solution would apply to Calcium and Magnesium, whose combined concentration is ---- - HCO3 5.20 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- 2- anions SO4 2.60 Cl- 2.51 5.20 5.80 10.31

Find: Mg S O4 mg (hypothetical) L Hardenss=6.70 Ion C Ca2+ 4.89
meq Hardenss=6.70 meq L L Ion C Ca2+ 4.89 cations Mg2+ 1.81 4.89 6.70 10.31 Na+ 3.61 6.70 milliequivalents per liter. To convert these units of calcium carbonate, ---- - HCO3 5.20 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- 2- anions SO4 2.60 Cl- 2.51 5.20 5.80 10.31

Find: Mg S O4 mg (hypothetical) L Hardenss=6.70 Ion C x 50.1 Ca2+ 4.89
meq Hardenss=6.70 meq L L Ion C mg equivalent x 50.1 weight meq Ca2+ 4.89 ofCaCO3 cations Mg2+ 1.81 4.89 6.70 10.31 Na+ 3.61 we’ll multiply by the equivalent weight of milligrams per milliequivalent, and the hardness would equal, ---- - HCO3 5.20 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- 2- anions SO4 2.60 Cl- 2.51 5.20 5.80 10.31

Find: Mg S O4 mg (hypothetical) L Hardenss=6.70 Ion C x 50.1 Ca2+ 4.89
meq Hardenss=6.70 meq L L Ion C mg equivalent x 50.1 weight meq Ca2+ 4.89 CaCO3 335.7 [mg/L] cations Mg2+ 1.81 4.89 6.70 10.31 Na+ 3.61 335.7 milligrams per liter, as calcium carbonate. - HCO3 5.20 SO4 Ca2+ Mg2+ Na+ HCO3 2- Cl- 2- anions SO4 2.60 Cl- 2.51 5.20 5.80 10.31

? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4

Find: Mg S O4 mg (hypothetical) L Ca2+ SO4 Mg2+ Na+ - HCO3 Cl- Ion C

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Find: Mg S O4 mg (hypothetical) L Ca2+ SO4 Mg2+ Na+ - HCO3 Cl- Ion C

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