⑴ 硫酸钙溶解度是什么
硫酸钙属微溶物质，也就是在常温下， 100g水中最多溶解的溶质质量在0.01g~~1g之间。如果是溶液中出来的微溶物质就加↓ 符号，如果反应物中已经有固体了，产生 的固体就不加 ↓ 符号，因此 CaO +H2O =Ca(OH)2 前面 CaO 是固体SO3 +Ca(OH)2=CaSO4 ↓+ H2O， 前面 Ca（OH）2是溶液形式，石灰水嘛
⑵ 硫酸钙溶解度与PH值的关系具体在PH小于7和大于7
PH小于7时,硫酸钙溶解度随PH值的增大而减小.CaSO4+2H+=Ca2++HSO4-PH大于7时,硫酸钙溶解度随PH值的增大而减小.CaSO4+2OH-=Ca（OH）2+SO42-（氢氧化钙的溶解度更小）
⑶ 硫酸钙溶度积的测定
难溶强电解质溶度积常数Ksp的测定一、 实验目的1、 了解极稀溶液浓度的版测量方法;2、 了解测定权难溶盐Ksp的方法;3、 巩固活度、活度系数、浓度的概念及相关关系。二、 实验原理 在一定温度下，一种难溶盐电解质的饱和溶液在溶液中形成一种多项离子平衡，一般表示式为:这个平衡常数Ksp称为溶度积常数，或简称溶度积，严格地讲Ksp应为相应个离子活度的乘积，因为溶液中个离子有牵制的作用，但考虑的难容电解质饱和溶液中离子强度很小，可警世的用浓度来代替活度。就AgCl而言 从上式可知，若测出难溶电解质饱和溶液中个离子的浓度，就可以计算出溶度积Ksp。因此测量最终还是测量离子浓度的问题。若设计出一种测量浓度的方法，就找到了测量Ksp的方法。具体测量浓度的方法，包括滴定法(如AgCl溶度积的测定)，离子交换法(如CuSO4溶度积的测定)，电导法(如AgCl溶度积的测定)，离子电极法(如氯化铅溶度积的测定)，电极电势法(Ksp与电极电势的关系)，即分光光度法(如碘酸铜溶度积的测定)等
⑷ 硫酸钙 溶解度
CaSO4溶解度不大，其溶解度呈特殊的先升高后降低状况。如10℃溶解度为0.1928g／100g水(下同)，40℃为0.2097，100℃降至0.1619。
⑸ 硫酸钙溶解度与PH值的关系
PH小于7时，硫酸钙溶解度随PH值的增大而减小。
CaSO4+2H+=Ca2++HSO4-
PH大于7时，硫酸钙溶解度随PH值的增大而减小。
CaSO4+2OH-=Ca（OH）2+SO42-（氢氧化钙的溶解度更小）
⑹ 硫酸钙溶解度和溶度积
溶解后的硫酸钙一部分以未电离硫酸钙存在，一部分电离成硫酸根离子和钙离子
所以专硫酸钙溶解度=钙或属硫酸根离子浓度+溶解态的硫酸钙浓度
硫酸钙溶解度2g/l，即0.015mol/L
离子对解离方程
CaSO4（aq）== Ca2+（aq） + SO42-（aq）
[Ca2+][SO42-]/[CaSO4]=5.2×10^-3
设钙离子浓度为x mol/L（等于硫酸根离子浓度），离子对态的硫酸钙浓度就是0.015-x mol/L，带入上面平衡方程：
x*x/（0.015-x）=5.2×10^-3
解得：x=6.6×10^-3
溶度积=[Ca2+][SO42-]=x*x=4.36×10^-5
我网上查了一下，硫酸钙溶度积相差比较大，有-7数量级也有-6数量级也有-5数量级的。我估计应该是-5数量级较合理一些，因为是微溶的。
另外，我根据热力学数据，推算出溶度积是7.1×10^-5
给你一个参考网站，溶度积与计算结果较为接近
⑺ 为什么硫酸钙饱和溶液要在干过滤以后才能用(用离子交换树脂除去钙离子之前)
因为半水以及无水硫酸钙本身具有吸水性。
纯水中加入过量的固回体，经常搅拌，恒答定在一定的温度下足够长时间，分别制备不同温度下的饱和溶液。吸取一定体积的饱和溶液（经干过滤后的），加入到装有阳离子交换树脂的离子交换柱中。收集经充分交换和洗涤后的溶液，准确测定pH或H+离子的物质的量（可用滴定法），可进而计算硫酸钙的溶解度。
钙离子浓度测定，也可以采用其它测定方法。
至于对应的固体含几个结晶水，你该明白做法。
至于不同水合物的溶解度，结合数据可进行理论推算。
⑻ 硫酸钙溶解度测定方法具体是什么啊
Solubility of CaSO4
Experiment 2
Major Concepts and Learning Goals
∙ Application of the solubility proct constant (Ksp)
∙ Saturated solutions
∙ Le Chatlier’s Principle/Common ion effect
∙ Activities and activity coefficients
∙ Ion selective electrodes
∙ Calibration curves
Laboratory Task
∙ Proce a calibration curve using standard solutions of CaNO3
∙ Measure the activities of the calcium ion, ACa2+, of three different solutions
1) a saturated solution of CaSO4 in H2O
2) a saturated solution of CaSO4 in 0.10 M KNO3
3) a saturated solution of CaSO4 in 0.10 M Na2SO4
∙ Use concepts of ionic strength () and activity coefficient () to calculate the concentrations and activities of Ca2+ and SO42- in each of the three solutions
∙ Calculate Ksp for CaSO4 using the data from each of the solutions and compare it to the literature value of 2.4•10-5.
Introction
The solubility of CaSO4 at 25 ºC is described by the following reaction and equilibrium
CaSO4(s) ↔ Ca2+ + SO42-
Ksp(CaSO4) = [Ca2+][SO42-] = 2.4∙10-5 Eq. 1
In words, this equilibrium expression states that the proct of the calcium ion concentration and the sulfate ion concentration can be no larger than 2.4∙10-5 in any aqueous solution.
Saturated solutions
Any aqueous solution in which the proct of the calcium ion concentration and the sulfate ion concentration is 2.4∙10-5 is said to be a saturated CaSO4 solution.
If a little more Ca2+ or SO42- is added to a saturated CaSO4 solution the equilibrium will shift to the left to form solid CaSO4 and the value of the proct of the calcium ion concentration and the sulfate ion concentration would be restored to 2.4∙10-5. This statement is the basis of Le Chatlier’s Principle.
When an equilibrium position of a reaction is disturbed, a new equilibrium position will be established by shifting the reaction in a direction that alleviates the stress caused by the disturbance
Saturated solutions can be prepared by a variety of methods. In this experiment the first saturated solution has been prepared by adding solid CaSO4 to purified water (the water comes from a purification system that includes a carbon filter, an ion-exchange resin and a UV lamp). The solution was mixed for several days and allowed to settle and reach equilibrium for several weeks. Because the only source of the calcium ions and sulfate ions are from the dissolution of CaSO4, [Ca2+] = [SO42-]. In fact, the same statement can be made for the second saturated solution, since KNO3 is not a source of Ca2+ or SO42-
Based on the literature Ksp value, and ignoring activities (see below), the [Ca2+] of these first two saturated solutions are about 5.0∙10-3 M.
Le Chatlier’s Principle and the Common Ion Effect
One general case in which Le Chatlier’s principle can be applied is when the solution contains a soluble salt of an ion that is in common with the insoluble salt in question. This is the case in the third saturated solution; CaSO4 in dissolved 0.10 M Na2SO4. In this solution there are two sources of the sulfate ion; the Na2SO4 and the CaSO4. The [SO42-] concentration coming from the Na2SO4 is 0.10 M. The sulfate ion coming from the CaSO4 is equal to the calcium ion concentration.
Thus,
Ksp = [Ca2+] ([SO42-]CaSO4+ + [SO42-]Na2SO4) = 2.4∙10-5
Letting [Ca2+] = x, we arrive at
Ksp = x (x + 0.10) = 2.4∙10-5
If we assume x <<< 0.10 M, then
Ksp = x 0.10 = 2.4∙10-5 and [Ca2+] = 2.4∙10-4 M,
This is considerably lower than the first saturated solution. It is also important to note that our solution verifies that our stated assumption was valid.
This is lowering of the [Ca2+] is said to be e to the common ion effect.
Activities and Activity Coefficients
In reality, equilibria are affected by the concentration of ions, any ions, in solution. The ionic strength () is used as a measure of the total ion concentration of a solution. It is calculated by incorporating each of the “i” ionic species in solution to the following equation.
= ½ ∑(CiZi2)
So, why does the ionic strength matter? Well let’s look at the CaSO4 equilibrium as an example. The ionic strengths of the saturated solution 2 and 3 are considerably greater than that of saturated solution 1.
Saturated solution 1
= ½ ([Ca2+](2)2 + [SO42-](-2)2))
= ½ (5∙10-3(2)2 + 5.0∙10-3(-2)2)) = 0.02 M
Saturated solution 2
= ½ ([K+](1)2 + [NO3-](-1)2 + [Ca2+](2)2 + [SO42-](-2)2))
= ½ (0.1(1)2 + 0.1(-1)2 + 5∙10-3(2)2 + 5.0∙10-3(2)2)) = 0.12 M
Saturated solution 3
= ½ ([Na+](1)2 + [SO42-](-2)2) + [Ca2+](2)2)
= ½ (0.2(1)2 + 0.1(-2)2 + 2.4∙10-4(2)2 ) = 0.20 M
Electrostatic interactions between the negative ions and the Ca2+ ions in solution cause the formation of an ion cloud around the Ca2+ ions. The larger the ionic strength of the solution the greater the radius of this ion cloud. A similar ion cloud forms around the SO42- ions from the positive ions in solution. The size of the ion clouds about Ca2+ and SO42- defines the energetics associated with these ions finding each other in solution and forming CaSO4(s). Thus, the proct of the [Ca2+] and the [SO42-] in the second saturated solution would be expected to be greater than 5.0∙10-3 M, and in the third saturated solution the [Ca2+] would be expected to be greater than 2.4∙10-5 M and [SO42-] would be expected to be greater than 0.10 M.
Mathematically, the effect of ionic strength is accounted for by introcing the concept of ion activity, A. The activity of an ion can be thought of as its effective concentration and is given by proct of its concentration and activity coefficient, . The activity coefficient depends upon the size of the ion, its charge, and the ionic strength of the solution. For example:
ACa2+ = [Ca2+] Ca+
The activity coefficient can be calculated using the Debye-Huckle equation.
In this experiment estimates of the activity coefficients of Ca2+ and SO42- for the three saturated solutions will be provided with the corresponding spreadsheet.
The most important concept to appreciate in terms of activities is that the definition of equilibrium expression that we first learn and used is a model that works very well at low ionic strengths, where the activity coefficients are close to unity. But the more universal model expresses the equilibrium constants in terms of activities. Equation 1 becomes
Ksp(CaSO4) = ACa2+ASO42- = [Ca2+]Ca2+[SO42-]SO42- = 2.4∙10-5 Eq. 2
Ion-Selective Electrodes
Ion-selective electrodes are electrochemical cells that have been carefully crafted into a probe that is useful for measuring the activity of a specific ion in solution. An ion-selective electrode most often consists of two reference electrodes, which give a constant potential. The two reference electrodes are electrically linked via a voltmeter and a salt bridge through the solution being measured. The salt bridge consists of a membrane between the solution being measured and the inner reference electrode solution. This membrane is made of a unique material that preferentially allows the ion of interest to partially penetrate the membrane. This partial penetration leads to the development of a junction potential, the magnitude of which depends entirely on the activity of the ion in the solution being measured. The Ca2+ selective electrode used in this application uses a membrane that preferentially chelates Ca2+. The potential measured by the voltmeter depends upon the potential at each of the reference electrodes and the junction potential. The potential of the reference electrodes is constant and junction potential is only influenced by the activity of the Ca2+ in the solution being measure.
Ecell = constant + 59.16/2 log ACa2+
Typically, one would measure the voltage of several standard Ca2+ solutions of known activities in the range of 0.1-10-6 M. Then, a calibration plot is proced by plotting the cell potential as a function of the log ACa2+.
In the Lab
1. Make serial dilutions of the 0.100 M Ca(NO3)2 solution using a 10.00 mL pipet and several 100 mL volumetric flasks. You should end up with solutions at the following concentrations; 0.0100 M, 1.00∙10-3 M, 1.00∙10-4 M, 1.00∙10-5 M. The Corresponding activities of these solutions are as follows 0.0100 M, 1.00∙10-3 M, 1.00∙10-4 M, 1.00∙10-5 M.
2. Carefully pour about 25 mL of the three CaSO4 solutions into each of three 100 mL beakers. Label the beakers well.
3. Measure the cell voltages of each of the six standards and the three samples of different saturated CaSO4 solutions. Record all data on the Lab Pro Software and and paste the data into an Excel spread sheet.
Lab Report
Open the Excel file named “CASO4”. In the “standard curve” worksheet enter the Ecell readings. Be sure that the cells where the data are inputted correspond to the correct concentration of the standard. The graph will show the calibration plot (mV vs log ACa2+) for your Ca2+ ion selective electrode. The slope and intercept of the plot is also given and automatically placed on top of the worksheet labeled “CaSO4 solutions”. Go to this worksheet. Place the data you obtained for the cell potentials for your three CaSO4 solutions. For the first and second saturated solutions (saturated CaSO4 in H2O and saturated CaSO4 in 0.1 M KNO3) the spreadsheet calculates the ACa2+ from the standard curve parameters, then calculates the [Ca2+] using the estimated activity coefficient for Ca2+, which, as described above, must equal the [SO42-], then calculates the ASO42- from the estimated activity coefficient of SO42-, and finally we calculated the Ksp of CaSO4 from the proct of the activities.
log ACa2+ = (Ecell – intercept)/slope
ACa2+ = 10 (log ACa2+)
[Ca2+] = ACa2+ / Ca2+
[SO42-] = [Ca2+]
ASO42- = [SO42-] SO42-
Ksp(CaSO4) = ACa2+ ASO42-
For the saturated CaSO4 solution in 0.10 M Na2SO4 the spreadsheet performs the same calculations, except that the [SO42-] is assumed to be 0.10 M.
The average Ksp of CaSO4 is also calculated.
Questions:
1. Compare and rationalize the relative magnitudes of the [Ca2+] in the three saturated CaSO4 solutions.
2. Are the three values for the Ksp of CaSO4 calculated from your experimental data in reasonable agreement?
3. Is the average value for the Ksp of CaSO4 calculated from your experimental data in reasonable agreement with the literature value of 2.4∙10-5?
⑼ 硫酸钙的常温下溶解度大概在0.27-0.4g/100ml，是微溶还是难溶
溶解后的硫酸钙一部分以未电离硫酸钙存在，一部分电离成硫酸根离子和钙离子
所以硫酸钙溶解度=钙或硫酸根离子浓度+溶解态的硫酸钙浓度
硫酸钙溶解度2g/l，即0.mol/L
离子对解离方程
CaSO4(a)== Ca2+(a) + SO42-(a)
[Ca2+][SO42-]/[CaSO4]=5.2×10^-3
设钙离子浓度为x mol/L(等于硫酸根离子浓度)，离子对态的硫酸钙浓度就是0.-x mol/L，带入上面平衡方程:
x*x/(0.-x)=5.2×10^-3
解得:x=6.6×10^-3
溶度积=[Ca2+][SO42-]=x*x=4.36×10^-5
我网上查了一下，硫酸钙溶度积相差比较大，有-7数量级也有-6数量级也有-5数量级的。我估计应该是-5数量级较合理一些，因为是微溶的。
另外，我根据热力学数据，推算出溶度积是7.1×10^-5
给你一个参考，溶度积与计算结果较为接近