⑴ 硫酸鈣溶解度是什麼
硫酸鈣屬微溶物質，也就是在常溫下， 100g水中最多溶解的溶質質量在0.01g~~1g之間。如果是溶液中出來的微溶物質就加↓ 符號，如果反應物中已經有固體了，產生 的固體就不加 ↓ 符號，因此 CaO +H2O =Ca(OH)2 前面 CaO 是固體SO3 +Ca(OH)2=CaSO4 ↓+ H2O， 前面 Ca（OH）2是溶液形式，石灰水嘛
⑵ 硫酸鈣溶解度與PH值的關系具體在PH小於7和大於7
PH小於7時,硫酸鈣溶解度隨PH值的增大而減小.CaSO4+2H+=Ca2++HSO4-PH大於7時,硫酸鈣溶解度隨PH值的增大而減小.CaSO4+2OH-=Ca（OH）2+SO42-（氫氧化鈣的溶解度更小）
⑶ 硫酸鈣溶度積的測定
難溶強電解質溶度積常數Ksp的測定一、 實驗目的1、 了解極稀溶液濃度的版測量方法;2、 了解測定權難溶鹽Ksp的方法;3、 鞏固活度、活度系數、濃度的概念及相關關系。二、 實驗原理 在一定溫度下，一種難溶鹽電解質的飽和溶液在溶液中形成一種多項離子平衡，一般表示式為:這個平衡常數Ksp稱為溶度積常數，或簡稱溶度積，嚴格地講Ksp應為相應個離子活度的乘積，因為溶液中個離子有牽制的作用，但考慮的難容電解質飽和溶液中離子強度很小，可警世的用濃度來代替活度。就AgCl而言 從上式可知，若測出難溶電解質飽和溶液中個離子的濃度，就可以計算出溶度積Ksp。因此測量最終還是測量離子濃度的問題。若設計出一種測量濃度的方法，就找到了測量Ksp的方法。具體測量濃度的方法，包括滴定法(如AgCl溶度積的測定)，離子交換法(如CuSO4溶度積的測定)，電導法(如AgCl溶度積的測定)，離子電極法(如氯化鉛溶度積的測定)，電極電勢法(Ksp與電極電勢的關系)，即分光光度法(如碘酸銅溶度積的測定)等
⑷ 硫酸鈣 溶解度
CaSO4溶解度不大，其溶解度呈特殊的先升高後降低狀況。如10℃溶解度為0.1928g／100g水(下同)，40℃為0.2097，100℃降至0.1619。
⑸ 硫酸鈣溶解度與PH值的關系
PH小於7時，硫酸鈣溶解度隨PH值的增大而減小。
CaSO4+2H+=Ca2++HSO4-
PH大於7時，硫酸鈣溶解度隨PH值的增大而減小。
CaSO4+2OH-=Ca（OH）2+SO42-（氫氧化鈣的溶解度更小）
⑹ 硫酸鈣溶解度和溶度積
溶解後的硫酸鈣一部分以未電離硫酸鈣存在，一部分電離成硫酸根離子和鈣離子
所以專硫酸鈣溶解度=鈣或屬硫酸根離子濃度+溶解態的硫酸鈣濃度
硫酸鈣溶解度2g/l，即0.015mol/L
離子對解離方程
CaSO4（aq）== Ca2+（aq） + SO42-（aq）
[Ca2+][SO42-]/[CaSO4]=5.2×10^-3
設鈣離子濃度為x mol/L（等於硫酸根離子濃度），離子對態的硫酸鈣濃度就是0.015-x mol/L，帶入上面平衡方程：
x*x/（0.015-x）=5.2×10^-3
解得：x=6.6×10^-3
溶度積=[Ca2+][SO42-]=x*x=4.36×10^-5
我網上查了一下，硫酸鈣溶度積相差比較大，有-7數量級也有-6數量級也有-5數量級的。我估計應該是-5數量級較合理一些，因為是微溶的。
另外，我根據熱力學數據，推算出溶度積是7.1×10^-5
給你一個參考網站，溶度積與計算結果較為接近
⑺ 為什麼硫酸鈣飽和溶液要在干過濾以後才能用(用離子交換樹脂除去鈣離子之前)
因為半水以及無水硫酸鈣本身具有吸水性。
純水中加入過量的固回體，經常攪拌，恆答定在一定的溫度下足夠長時間，分別制備不同溫度下的飽和溶液。吸取一定體積的飽和溶液（經干過濾後的），加入到裝有陽離子交換樹脂的離子交換柱中。收集經充分交換和洗滌後的溶液，准確測定pH或H+離子的物質的量（可用滴定法），可進而計算硫酸鈣的溶解度。
鈣離子濃度測定，也可以採用其它測定方法。
至於對應的固體含幾個結晶水，你該明白做法。
至於不同水合物的溶解度，結合數據可進行理論推算。
⑻ 硫酸鈣溶解度測定方法具體是什麼啊
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 硫酸鈣的常溫下溶解度大概在0.27-0.4g/100ml，是微溶還是難溶
溶解後的硫酸鈣一部分以未電離硫酸鈣存在，一部分電離成硫酸根離子和鈣離子
所以硫酸鈣溶解度=鈣或硫酸根離子濃度+溶解態的硫酸鈣濃度
硫酸鈣溶解度2g/l，即0.mol/L
離子對解離方程
CaSO4(a)== Ca2+(a) + SO42-(a)
[Ca2+][SO42-]/[CaSO4]=5.2×10^-3
設鈣離子濃度為x mol/L(等於硫酸根離子濃度)，離子對態的硫酸鈣濃度就是0.-x mol/L，帶入上面平衡方程:
x*x/(0.-x)=5.2×10^-3
解得:x=6.6×10^-3
溶度積=[Ca2+][SO42-]=x*x=4.36×10^-5
我網上查了一下，硫酸鈣溶度積相差比較大，有-7數量級也有-6數量級也有-5數量級的。我估計應該是-5數量級較合理一些，因為是微溶的。
另外，我根據熱力學數據，推算出溶度積是7.1×10^-5
給你一個參考，溶度積與計算結果較為接近