1、1,Hydrogeochemistry 水文地球化学 By Prof. Dr. Sun Zhanxue,2,Chapter 1 Introduction and chemical background 引言与化学背景,3,LEARNING OBJECTIVES,Review basic fundamentals of chemistry. Understand commonly used concentration units and unit conversions. Learn to calculate and understand the significance of water ha
2、rdness and charge-balance error. Be introduced to some common ways of graphically displaying natural water compositions.,4,1.1 Introduction(引言),5,Scope of aquatic chemistryAquatic chemistry is concerned with the chemical reactions and processes affecting the distribution and circulation of chemical
3、species in natural waters. The objectives include the development of a theoretical basis for the chemical behavior of ocean waters, estuaries(河口、海湾), rivers, lakes, groundwaters, and soil water systems, as well as the description of processes involved in water technology. W. Stumn & J.J. Morgan (199
4、6）,6,Water chemistryWater chemistry has a potential use for tracing the origin and the history of water. Water compositions change through reactions with environment, and water quality may yield information about environments through which the water has circulated. There is a great need in present-
5、day hydrology to obtain better knowledge concerning residence times, flowpaths and aquifer characteristics, a need which often arises from man-made pollution problems. The water chemistry can provide the required information since chemical reactions are time and space dependent.,7,1.1.1 水文地球化学的发 展 历
6、 史（Histories of hydrogeochemistry),国外: 古希腊思想家亚里斯多德指出：“流经怎样的岩石就有怎样的水”。 古罗马，开始广泛利用地下热水，并研究其成分，将矿水分为碱性矿水、铁质矿水、含盐矿水和含硫矿水。,8,18世纪中页，罗蒙诺索夫的论地层指出天然水是一种复杂的溶液。 20世纪初，维尔纳茨基为水化学及水文地球化学的奠基人，C.A.舒卡列夫提出水化学分类。 1921年苏联科学院成立了世界上第一个水文化学研究所,9,国外主要的水化学方面的著作,1948年 O.A.阿列金 普通水文化学 1958年 A.U.别列尔曼 景观地球化学概论 1965年 R.M.Garrels
7、 Solutions Minerals and Equilibria 1969年 K.E.比契叶娃 地下水区域地球化学基础 1970年 O.A.阿列金 水文化学基础 1970年 A.M.奥弗琴尼柯夫 水文地球化学 1975年 A.U.别列尔曼 后生地球化学 1977年 B.C.萨玛丽娜 水文地球化学 1978年 K.E.比契叶娃 水文地球化学 1978年 A.K.利西秦 成矿作用水文地球化学,10,1979年 H.沃兹娜娅 水化学与微生物学 1982年G. Matthess The Properties of Groundwater 1988年 J.I. Drever The Geochemi
8、stry of Natural Waters（第二版）（1997已出第三版） 1988年 R.M. Pytkowicz 平衡非平衡与天然水 1989年 J.D. Hem Study and interpretation of the chemicalcharacteristics of natural water 1994年 C.A.J. Appelo & D. Postma Geochemistry, Groundwater and Pollution 1996年 W. Stumm & J.J. Morgan Aquatic Chemistry 1997年 Langmuir, AQUEOUS
9、 ENVIRONMENT GEOCHEMISTRY 2001 Kehew, Applied chemical hydrogeology, Prentice-Hall,11, 国内: 我国是文明古国，四千多年前我们的祖先已凿井取水。 秦代 李冰，四川凿井求盐的创始人。 北宋 沈括(10311095)在梦溪笔谈中记载了铅山县用“苦泉”炼铜。 明代 李时珍(15191593)著本草纲目提出按泉水成分分成五类：(1)硫酸泉；(2)珠砂泉；(3)矾石泉；(4)雄黄泉；(5)砒石泉。 ,12,国内主要的水化学方面的著作,1926年 章鸿钊中国温泉辑要 1977年 水工所 水文地球化学找矿方法 1978年
10、水工所 中国地下水 1980年 高万林等 放射性水文地球化学找矿 1982年 李学礼 水文地球化学 1983年 沈照理 水文地球化学基础 1983年 张人权 同位素在水文地质学中的应用 1984年 李昌静、卫中鼎 地下水水质及其污染 1984年 刘存富 同位素水文地质学基础 1985年 水工所 水文地球化学理论与方法的研究,13,1986年 王秉忱 地下水污染地下水水质模拟方法 1988年 尹观 同位素水文地球化学 1988年 刘崇喜 水文地球化学找油理论与方法 1988年 李学礼 水文地球化学(第二版) 1990年 史维浚 铀水文地球化学 1993年 沈照理 水文地球化学基础(新版) 199
11、3年 李宽良 水文地球化学热力学 1993年 李雨新 水溶液理论概论 ,14,天然水的组成：海洋：96%冰：3%地下水: 1% 溪流、湖泊：0.01%大气: 0.001%,滞留时间的概念T = 总水量 / 输入水量河水和地下水排入海中的水量为：0.361020g/y降水补充海水的水量为： 3.51020g/y。故：,水 循 环,1.1.2 水文学中的某些基本概念Some concepts in hydrology（pp.1-13),15,16,雨水的化学成分和水文地质中的化学术语 (I),雨水不是化学纯水 近海地区：含海盐，Na, Cl 含量高；远离海洋地区：含Ca , SO4较高。雨水中还含
12、有甲酸（Formic Acid）、乙酸（Acetic Acid)、NOx 、 SO2 等。,水文地质学中的化学术语溶解固体总量（TDS-Total Dissolved Solids）: 溶解组分总和 HCO3/2 (等同于矿化度）盐度（Salinity): 溶解组分总和.硬度（Hardness）：水中与钠皂反应生成沉淀物的离子浓度。以CaCO3 表示：H(mg/l)=2.5Ca +4.1 Mg,17,雨水的化学成分和水文地质中的化学术语(II),淡水（Fresh water): TDS 1000mg/L 微咸水（Brackish water): 1000mg/LTDS2000mg/L 咸水（S
13、aline water): 2000mg/LTDS 35000mg/l,卤水（Brine):TDS 35000mg/l 注意：划分的数字标准并非绝对，仅是一般的分法。,18,1.2 化学背景(Chemical background) (pp.15-40),1.2.1 Aqueous solution and properties of water,19,20,STRUCTURE OF WATER,Ionic character: O - H = 3.5 - 2.1 = 1.4 39% ionic; 61% covalent - water is polar covalent. Water is
14、 a polar substance, meaning there is a positive and a negative pole to the molecule.,109,+,-,the water dipole,21,HYDROGEN BONDING,Because each water molecule has a positive and a negative end, these can attract one another to form a hydrogen bond.,Hydrogen bonding of water is the reason for the anom
15、alous behavior of water compared to similar substances.,22,ION HYDRATION,Also because of the polar nature of water, ions will be surrounded by water dipoles (hydrated) in solution.Hydration isolates the ions from their neighbors and neutralizes the attractive forces that hold minerals together.,23,I
16、n general, a relatively small number of inorganic constituents occur in substantial concentrations in most natural waters. Major constituents Ions: bicarbonate (HCO3-), calcium (Ca2+), chloride (Cl-), magnesium (Mg2+), sodium (Na+), and sulfate (SO42-). Neutral species: silica (H4SiO40) There are nu
17、merous possible minor and trace constituents.,COMPOSITION OF NATURAL WATERS,24,1.2.2 Concentration and unit conversion,Mass concentrations Water analyses are most commonly expressed in terms of the mass contained in a liter of solution (mg L-1, g L-1, ng L-1),25,Closely related to mg L-1 is parts pe
18、r million (ppm) or mg kg-1These two units are related through the density of the solution () or mass per unit volume.,26,Mass concentrations (continued) The conversion factor between mg L-1 and ppm is:Because the density of many natural waters is near 1 kg L-1, it is often a sufficiently good approx
19、imation that mg L-1 and ppm are numerically equal.,27,Mass concentrations (continued)Ambiguity can arise for some components of natural waters. For example, we can express the concentration of sulfate as mg L-1 SO42- or mg L-1 sulfate-S. The relationship among these is:,28,Molar concentrations In al
20、most all geochemical calculations, it is necessary to use molar concentrations rather than mass concentrations. Molarity (M) = moles of solute/liter of solution Molality (m) = moles of solute/1000 g of solvent If the density of the solution is significantly different from 1 kg L-1, then molality and
21、 molarity will be quite different; however, in most natural waters, these quantities are nearly equal and the difference between them can be neglected.,29,水溶液中溶质的浓度： m(molal units) 每千克溶剂中溶质的摩尔数(重量摩尔浓度)。M（Molar units) 每升溶液中溶质的浓度(体积摩尔浓度)。,30,Molar concentrations (continued) Conversion from mol L-1 (M)
22、 to mg L-1 is accomplished using the formula:where FW is the formula weight of the substance in g mol-1. The reverse conversion is accomplished using:,31,Mole fraction Another form of molar concentration, the mole fraction (X) is used for solid solutions, e.g., solid solutions between KAlSi3O8 and N
23、aAlSi3O8. In such a solid solution, the mole fraction of KAlSi3O8 would be written as:,32,固溶体的浓度一般用摩尔分数(mole fractions) 表示。A物质的摩尔分数等于某固溶体中A物质的摩尔数除以该固溶体中物质的摩尔总数。,33,Equivalents and Normality Equivalents (eq) are similar to moles, but take into account the valence of an ion. For example, 0.002 mol L-1
24、 of Ca2+ = 0.004 eq L-1 Ca2+; 0.001 mol L-1 of Na+ = 0.001 eq L-1 Na+; and 0.003 mol L-1 La3+ = 0.009 eq L-1 La3+. Normality (N) is another name for eq L-1. Alkalinity is an important solution parameter that is expressed as eq L-1 or meq L-1. Hardness is another parameter expressed as eq L-1.,34,在25
25、C,一个大气压下，一密度为1.020含2.00克CaCl2/L的溶液,问体积摩尔浓度、当量浓度和重量摩尔浓度？ 解：体积摩尔浓度2.00/MW(分子量)0.0045M,当量浓度=2体积摩尔浓度=0.009N 每升溶液的重量=1.02kg 其中溶剂的重量=1.020.002= 1.018kg 重量摩尔浓度= 0.0045/1.018 =0.00442m,例题1,35,例题2,在Ca-MgCO3固溶体中含有5wt%的Mg, 问MgCO3在该固溶体中的摩尔分数？解： Wt% MgCO3 =5(2460) / 24 = 17.5 Wt% CaCO3 = 10017.5 = 82.5,MgCO3 的相对
26、摩尔数 = 17.584 0.21 CaCO3的相对摩尔数= 82.5 100 0.825 MgCO3在该固溶体中的摩尔分数 =moles MgCO3 / mole MgCO3 moles CaCO3 = 0.21/(0.21+0.825)= 0.20,36,1.2.3 CHARGE-BALANCE ERROR CHARGE-BALANCE ERROR - I,Aqueous solutions must be electrically neutral. In other words, the sum of all negative charges must equal the sum of
27、all positive charges. One check on the quality of a water analysis is the charge-balance error, calculated as follows:,37,CHARGE-BALANCE ERROR - II,There is always some error in the measurement of cation and anion concentrations. Thus, we cannot expect a charge-balance error of zero for any analysis
28、. The C.B.E. may be positive or negative, depending on whether cations or anions are more abundant. A reasonable limit for accepting an analysis as valid is 5%.,38,REASONS FOR C.B.E. VALUES GREATER THAN 5%,An important anion or cation was not included in the analysis. Sometimes this can point out th
29、e presence of a high concentration of an unusual anion or cation. A serious, systematic error has occurred in the analysis. One or more of the concentrations was recorded incorrectly.,39,1.2.4 Graphical displays of water chemistry,40,PIPER DIAGRAMS - I,Consists of two triangles (one for cations and
30、one for anions), and a central diamond-shaped figure. Cations are plotted on the Ca-Mg-(Na + K) triangle as percentages. Anions are plotted on the HCO3-SO42-Cl- triangle as percentages. Concentrations are in meq L-1. Points on the anion and cation diagrams are projected upward to where they intersec
31、t on the diamond.,41,Figure 1-6 from Kehew (2001). Water analyses plotted on a Piper diagram. Cation percentages in meq L-1 plotted on the left triangle, and anion percentages in meq L-1 plotted on the right triangle.,42,PIPER DIAGRAMS - II,ADVANTAGES Many water analyses can be plotted on the same d
32、iagram. Can be used to classify waters. Can be used to identify mixing of waters. DISADVANTAGE Concentrations are renormalized. Cannot easily accommodate waters where other cations or anions may be significant.,43,Figure 1-7 from Kehew (2001). Classification of hydrochemical facies using the Piper p
33、lot.,44,An example of a Piper diagram drawn for mine waters from the Pine Creek district, Coeur dAlene Valley, ID. These may be characterized as Ca-Mg sulfate-bicarbonate-type waters.,45,TOTAL DISSOLVED SOLIDS,Total dissolved solids (TDS) - a useful property of an aqueous solution determined from th
34、e weight of the evaporated residue of a known volume of water. Roughly equal to the weight obtained by summing the masses of individual species in solution. Can be depicted on a Piper diagram using a circle centered on the data point with a circumference proportional to TDS.,46,An example of a Piper
35、 diagram with TDS circles.,47,TDS AND WATER CLASSIFICATION,Fresh waters - sufficiently dilute to be potable (TDS 1,000 mg L-1). Brackish waters - too saline to be potable, but significantly less saline than seawater (TDS = 1,000 - 20,000 mg L-1). Saline waters - similar to or more saline than seawat
36、er (TDS 35,000 mg L-1). Brines - significantly more saline than seawater.,48,49,1.2.5 Equilibrium thermodynamics (平衡热力学）,In natural systems, complete chemical equilibrium is rarely attained, particularly where biological process are involved. Nevertheless, the equilibrium approach is useful in that
37、(1) it often provides a good approximation to the real world,50,(2) it indicates the direction in which changes can take place (in the absence of some energy input, systems, including biological systems, can move only toward equilibrium), and,51,(3) it is the basis for calculation of rates of natura
38、l processes since, in general, the farther a system is from equilibrium, the more rapidly it will react toward equilibrium, although it is rarely possible to predict reaction rates quantitatively from thermodynamic data.,52,A system at equilibrium is characterized by being in a state of minimum ener
39、gy. A system not at equilibrium can move toward equilibrium by releasing energy. For system at constant temperature (T) and pressure (P), the appropriate measure of energy is the Gibbs free energy (G), which is related to enthalpy (heat content)(热焓) H and entropy（熵） S by the equation G = H TS,53,The
40、 units of G and H are commonly kJmol-1; the unit of S is Jmol-1K-1. T is the temperature on the Kelvin scale. For changes at constant P and T,G = H T S Equilibrium constantaA bB = cC dDThe Gibbs free energy per mole of reaction, G R=GproductsGreactants,54,where,G is the standard free energy of react
41、ion, that is, the change in free energy when a mole of A and b moles of B are converted to c moles of C and d moles of D, all in their standard states. At equilibrium, G =0 and hence:,55,Where Keq is the equilibrium constant.,56,The equilibrium constant is thus related to the stand free energy of re
42、action by equationor,57,For temperature not far from 25C, the variation of Keq with temperature can be calculated from the equationsG = H T S= -RTLn Keq which rearranges toValues of HR and SR can be obtained from the standard-state enthalpies and entropies of formation of reactants and products.,58,
43、This method of calculation is based on the assumption that HR and SR are independent of temperature. The assumption is reasonable for temperatures with 20C or so of the standard state.An alternative method based on the same assumptions come from the differentiating equation,59,and integrating this f
44、rom T1 (the reference temperature) to T2 (the temperature of interest):Thus, if the equilibrium constant at one temperature and the standard enthalpy of reaction are known, the equilibrium constant at another temperature can readily be calculated. If T2 is far from T1,60,the variation of HR with tem
45、perature must be calculated from heat capacity data.,61,Measurements of Disequilibrium,Consider again the reaction aA bB = cC dDIf the system is at equilibrium,62,If the system is not at equilibrium, the expression on the left side of the above Equation, called the activity products (AP) or the ion
46、activity product (IAP) if the species involved are ions, will not be equal to Keq. If AP/ Keq1, the reaction will tend to go to the left; if AP/ Keq1, the reaction will tend to go to the right. Expressions used by various authors to indicate how far a system is from equilibrium include AP/ Keq (whic
47、h,63,Will be 1 at equilibrium), log(AP/Keq) (which will be zero at equilibrium), and RT ln(AP/Keq).RT ln(AP/Keq) it represents the free- energy change in converting 1 mol of reactants to products under the conditions at which the AP was measured. Saturation IndexSI= log(IAP/Ksp),64,SI0 Supersaturate
48、da particular solution with respect to a particular mineral.What is SR? (p.25),65,Example 3,A water has a calcium activity of 10-3.5 and a sulfate activity of 10-1.5. By how much is it undersaturated (or super-saturated) with respect to gypsum? By how much would it have to concentrated (or diluted)
49、for it to be in equilibrium with gypsum?,66,IAP 10-3.5 10-1.5 10-5.0 Ksp = 10-4.36 SI=Log(IAP/Ksp) = Log0.23 = -0.64 0 undersaturated,67,Question?,IAP/Ksp=0.231 The solution would have to be concentrated by a factor of ? to achieve equilibrium with gypsum.?=1/0.23=4.35 Or?=(4.35)1/2=2.09 Correct,68,
50、RTLn(IAP/Ksp)=3.65This last number indicates that, if gypsum were to dissolve in water in question. 3.65 kJ of energy would be released per mole of gypsum dissolved. This, of course, is the instantaneous rate at which energy would be released. As gypsum dissolved, the activity of Ca2+ and SO42- would increase, and the energy released per mole of dissolved gypsum would decrease until it became zero when the solution became saturated with respect to gypsum.,
