Solubility Product Constant Lab Report

In chemical reactions, the significance of knowing the limiting reactant is high. In order to increase the percent yield of product, increasing the limiting reactant, possibly, is the most effective. In this experiment we were able to calculate limiting reactants from the reaction of CaCl2. 2H2O + K2C2O4.H2O(aq).

In this experiment, I was trying to determine the solubility product constant (Ksp) of Lead(II) Chloride. This experiment is important because it will allow me to determine the concentration of ions in an aqueous state from a solute such as Lead(II) Chloride.

I believe that when the water temperature increases the dissolving rate would increase whereas when the temperature of the water decreases the dissolving rate would decrease.

A 300.0-mL saturated solution of copper(II) peroidate (Cu(IO4)2) contains 0.38 grams of dissolved salt. Determine the Ksp.

Purpose: The purpose of this lab is to use the three equations of solution chemistry to determine the molarity and percent composition by weight of a solution.

After the calculations were complete the correct amounts of the reactants had to be obtained. 75-100 ml of water was added to the 3 grams of Ca(CO3). Then the 30 ml of H2(SO4) was added slowly to the Ca(CO3) solution. 20ml more of water was added then the reactants sat for five minutes. Once they sat for five minutes, the precipitate was filtered through filter paper. After the precipitate had time to dry,it was weighed.

The hypothesis stated the higher temperature the water is, the faster the sugar cube will dissolve. The reference tables and graph show that the warmer the water is, the faster the sugar cube will dissolve. The reference tables and graph also show that the colder the water is the bigger the difference between the rates of dissolving are. For example the sugar dissolving in the 5.0 (℃) water took almost 8 minutes longer to dissolve compared to the sugar in the 88 (℃) water. The warmer the water became, the smaller the difference between the rates of dissolving were. Figure 1 shows the average dissolving rate at different temps.

The purpose of this lab is to learn how to identify a solution as unsaturated, saturated or supersaturated. In this lab, we adjusted the amount of solute, the substance that does the dissolving, in the solvent, the substance that the solute dissolves in, to see the saturation levels.The amount of solute that a given amount of solvent will hold is called solubility. First, we added a crystal of sodium thiosulfate into 1 ml of water and observed that the crystal dissolved. This process indicates that the liquid was unsaturated which allowed or the solute to dissolve. Next, we added 2 grams of hypo powder. We observed that some of the powder dissolved but some didn’t and the test tub became cold. The coldness indicates

Calcium: Formula Unit: Ca(〖〖OH)〗_2〗_((aq) ) +〖K_2 C_2 O_4〗_((aq) ) →〖CaC_2 O_4〗_((s) ) +2KOH_((aq))

The hypothesis, “If the mass of solid CaCl2 and NH4Cl increases, then the qsolution (J) and ΔHsolution (kJ/mol) will increase proportionally.” is partially rejected. For CaCl2, the q solution does increase, but the ΔH is constant, not proportional. In the NH4Cl, the q solution decreases proportionally, while the ΔH is constant. In the calcium chloride reaction, the data is precise because the R squared value is .894 when the outlier is included which is relatively high considering the outlier. The ΔH data is relatively precise as well. The standard deviation of the ΔH without the outlier is only 3.38, but with the outlier the standard deviation rockets up to 15.9. Although the the data is precise, the accuracy is somewhat subpar. The percent error for ΔH is a strong 29.8%, which is far too high to be considered accurate.

Observe • To identify the molar solubility and the constant associated with it for calcium hydroxide. • To conduct an experiment regarding how the effect common ion has on the overall solubility of calcium hydroxide. Procedure 1. Molar Solubility and Solubility Product of Calcium Hydroxide • Obtain a prepared saturated solution containing calcium hydroxide.

Titration is a technique that has been used in this experiment to identify the Ksp value of calcium hydroxide in order to determine the extent to which the compound is soluble in water. A known volume of 50 mL of hydrochloric acid, a concentration of 0.05 M hydrochloric acid, a volume of 50 mL calcium hydroxide base, an unknown concentration of base and an acid-base indicator (bromothymol blue) has been used in this experiment to determine the concentration of calcium hydroxide. From the results tabulated, a Ksp value of 1.26 x 10-7 has been retrieved indicating that

In the first part of the experiment, the stock solution was prepared with a known initial volume, a known initial concentration, and a known final volume. The final concentration was calculated using the dilution equation, which showed how dilute or how concentrated the solution was as the experiment progressed. To form a solution at a certain concentration, such as a more diluted solution for an experiment, a solute is dissolved in a solvent thoroughly, and brought to the required volume of solution. Throughout the experiment, this was done repeatedly. In the first part of the experiment, as table 1 states, the final molarity concentration of test tube 1 was .10 M. As it progressed and the dilution equation was used to calculate the final concentrations of the new solutions, test tube 2 had a final molarity of .01, test tube 3 had .001 M, test tube 4 and 5 having .0001 M and .00001 M respectively. With each test tube, there was a progressively less concentrated initial molarity of the solution, which resulted in the less concentrated final molarities. For example, test tube 1 which began with an initial molarity of .10 M, and ended with a final molarity of .10 M, is significantly more concentrated than test tube 5, which had an initial molarity of .0001 and a final molarity of .00001. The results show that as the smaller molarity of initial solution was used, the smaller the final molarity was. This was also evident through the color change throughout the test tubes. The

We obtained a composition of 58 (±5.1) wt% Na2CO3 and 42 (±8.3) wt% NaHCO3 for the unknown solid analyzed in this experiment. Although the accuracy of these results cannot be fully evaluated, since we do not have a known composition concentration to compare, we must assume that there was some contribution of error, especially given the somewhat large errors observed. One possible source is a miscalculation of the volumes or concentrations obtained from the results due to exposure of CO2 in our titrants and solutions which, as discussed in the previous lab report, can make our solution more acidic than what the nature of the experiment requires.2 Therefore, different volumes of standard HCl would be necessary for titration. A simple solution for this issue would have been boiling the distilled water to expel carbon dioxide from the stock solutions. Additionally, the stock solutions may also be responsible for some of this error. Besides having stock solutions available for the entire class, which may cause cross contamination and mishandling of the stock solutions, the stock solution needed to be replenished multiple times, thus providing the possibility of preparing a solution with a different molarity. Moreover, at one point all 0.1 M NaOH had been expended, and the only option available was to use a solution of 0.08 M NaOH for further experiments. This was

The color changes, warming and effervescence was noted. The number of drops of solvent added to completely dissolve the sample was also noted. Based on the amount of solvent added, the sample was described as very soluble, soluble, slightly soluble, or insoluble. If the sample is liquid, the solubility was described as miscible, slightly miscible, or immiscible.