Determination of the Ksp of Calcium Hydroxide

**Abstract**

Solubility is the maximum amount of a solute to dissolve in a liquid (solvent), the solution is called saturated solution. One of example of solubility that keep aquatic species alive is the dissolvement of oxygen in water to help fishes and other species to breath. Various factors can affect solubility, under certain conditions the equilibrium solubility can be surpassed to give a supersaturated solution. There are three different methods to determine the solubility of a solution. In order to find the best method different conditions were applied to the solution and in each method the solubility product constant (Ksp) was found by using the molar solubility of hydroxide ion. In each method different concentration of the solution were used and the yield Ksp values were 4.12, 1.17, and 2.64respectively. Comparing these Ksp values to the known ksp of calcium hydroxide which is 4.68, they gave a percent error of 11.9%, 2400%, and 464.1% respectively. These results confirmed that the Ksp calculated by method was the closest to the theoretical Ksp value of calcium hydroxide. There are wide range of experiments to conduct to accomplish the lab goal, but by using method comparison we found that method 1 gave the best results.

**Introduction**

Solubility have a wide use in our lives, the easiest use of solubility is the dissolvement of sugar in our food. Solubility also have a negative impact on our lives, for example; the dissolving of toxic chemicals and metals in our water. In this experiment, method comparison is used to find the best experiment to determine the solubility equilibrium constant for calcium hydroxide, by using the method comparison we can determine the best method to mathematically calculate the Ksp value. Method one was expected to have the closest Ksp value to the theoretical one because solubility depends on temperature and heating the solution will increase its solubility.

In order to find the Ksp, first the molar solubility of the products need to be found experimentally. Calcium hydroxide is a saturated solution that is made by reacting solid sodium and water, then filtering the solid.

Ca(*s*) + H2O → Ca(OH)2(*s*) (1)

The test of solubility is used to determine how much of a solute is dissolved in the solvent under known conditions. One way to test the solubility of a solution is by applying temperature to the solution. As the temperature increase, the solubility of the solution increase too. When the temperature increase the movement of the solutes increases and this cause an increase in the average kinetic energy based on equation 1. As the kinetic energy of the solution increase the more effective the solvent molecules will become in breaking apart the solute molecules.In method 1 the filtered calcium hydroxide was heated to increase the solubility of the solid.

KE = 0.5 • mv2 (1)

m= mass

v= speed

Another way to test the solubility of a solution is by titration. In method three we titrated calcium hydroxide with hydrochloric acid, as we added HCl the concentration of hydrogen ion increased and the concentration of hydroxide ion decreased causing a right shift on reaction 2 based Le Chatelier’s principle, thus will cause a decrease in the solubility of Ca(OH)2. By finding the pH of the solution we can use equation two to mathematically find the concentration of hydroxide ion.

Ca(OH)2(*s*) Ca2+(*aq*) + 2OH−(*aq*) (2)

[OH] = 10 (2)

In method two and three we used equation number two to calculate the Ksp for each method. In method three we use the molar mass equation to find the moles of calcium hydroxide and then by using equation two we can find the Ksp.

Ksp = [X] [ Y] (3)

Molar mass = (4)

By rearranging equation 4, we can find the moles of the solution, after finding the moles of the solution we can find the molarity of the solution as shown in equation 6.

Moles = (5)

Molarity= (6)

After calculating the Ksp of each method then we can compare them to the theoretical Ksp value of calcium hydroxide and we can conclude which method was the best for the experiment.

**Data**

**Table1: Data table showing the collected resulted from method one. **

**Ca(OH)2 Volume (ml)** | **[Ca(OH)2] ** | **Ksp** |

50 ml | 0.01M | 4.12 |

**Table2: Data table highlighting the collected results from method two. **

**Ca(OH)2 Volume (ml)** | **pH** | **[OH]** | **Ksp** |

30 ml | 12.79 | 0.0617 M | 1.17 |

**Table3: The collected results from method three. **

**Ca(OH)2 Volume (ml)** | **HCl Volume** | **Methyl red indicator** | **[OH]** | **Ksp** | **Initial Color** | **Final Color** |

10 ml | 70 ml | 3 drops | 3.75 | 2.64 | Yellow | Light Pink |

**Result**

After collecting and filtering the calcium hydroxide in the Gravimetric Determination, it was placed in a beaker and then in an oven. Heating the solution will help to increase the rate of solubility. The beaker was then removed from the oven and set aside for couple seconds. By using equation 5, the moles of Ca(OH)2 was found and then the moles value was plotted in equation 6 to find the molarity of the solution, lastly, Ksp of the solution was calculated by using equation 3.

Moles = = 2.02

Molarity = = 0.01M **Ca(OH)2**

Ksp = 4x

Ksp = 4( 0.01)4.12

In method two the determination by pH, logger pro was used to record the pH of the filtered solution of calcium hydroxide. By finding the pH we can use equation 2 to calculate [OH]. Using the [OH] value as the molar solubility of the solution, the Ksp of the soluthen we have tresution can be determined by using equation 3.

[OH] = 10= 0.0617 [OH]

Ksp = (0.0614)= 1.17

Titration is the addition of an acid or a base to an acidic or basic solution to neutralize the solution. In this experiment a strong acid (HCl) was added to a strong base (**Ca(OH)2**), so the reaction will reach equilibrium when pH is approximately equal seven. By using the volume of HCl when the solution started to change color we can find the moles of hydroxide ions, and by dividing the moles by the total volume as shown in equation 6, the [OH] was found. Then by using equation 3 the Ksp can found.

7.5 ml = 3.75mol OH

Molarity = = 0.0375 [OH]

Ksp = (0.0375) = 2.64

Now knowing the ksp values for each method we can calculate the percent for each method as following:

Method 1 %error= = 11.9%

Method 2 %error = = 2400%

Method 3 $error = = 464.1%

**Table 4: Data table summarizing the result of the experiment. **

| **Method 1** | **Method 2** | **Method 3** |

**Ksp Values** | 4.12 | 1.17 | 2.64 |

**Percent error (%)** | 11.9% | 2400% | 464.1% |

As shown in table 4 and above calculation, method 1 had a ksp value of

4.12 which is the closest to the literature value which is 4.68. This indicate that temperature gave a big impact in the solubility of the solution.

**Discussion**

Determining the Ksp value of a solution could be hard, but by using method comparison, we can find the best fit procedure for the experiment. In this lab it was shown that there were three ways to determine the value of ksp. Each one of the method yield a different ksp value because each method was done under different conditions, for the first method the ksp value was determined by using temperature, the second method was by using the pH, and the third method was by using titration. Based on table 3 the first method had the lowest percent error which means it had the closest ksp value to the literature one, thus prove our hypothesis to be correct. Although each method yield a ksp value, they were problems with the experiment that could have affected the ksp value. One of the errors in the first method was the weight of the beaker, since the weight scale was not a closed system air could have affected the mass of the beaker, thus could have raised the ksp values. Another error was done in the titration method, the burette was rinsed with water instead of HCl and this caused the HCl that was later added to the burette to react with water and this reduced the strength of the aicd, thus caused the solubility of the solution to decrease. Having a closed system would have given better result, also better and more precise equipments could have yielded better result.

**Questions**:

A: The Ksp value would be lower than the true value because HCl would react with water and this will decrease the acidity of the acid.

B: The ksp value would be the same because the ng phenolphthalein has no effect on the ksp value it is just an indicator.

May 2015

**Thermodynamics of the Solubility of KNO3 LAB**

**Abstract: **

This experiment determined the Ksp, and some thermodynamic properties of potassium nitrate based off of its solubility. The results for temperature, Ksp, ΔH, ΔS, and ΔG were 341K-310K, 21.87M (avg), 24.92 kJ/mol (28.59% error), 0.1018 kJ/mol (12.24% error), and -8.253 kJ/mol respectively.

**Introduction:**

The purposed of the experiment was to determine the Ksp of the reaction and then use it to determine several thermodynamic properties. The reaction is likely to be endothermic making the reaction only spontaneous at a certain temperature.

The Ksp of potassium nitrate is given by the following reaction:

*Ksp *= [K+][NO3-] (0)

The thermodynamics of a reaction relate in these equations:

**ΔG = ΔH – TΔS (1)**

**ΔG = -RT***lnK*sp (2)

** (3)**

The experiment allowed the observing of the moment of first precipitation as the reaction proceeded in reverse. As ions first became solid again, the temperature was recorded and defined the temperature of the given Ksp value of each trial. After measuring the Ksp of the reaction (1), the given temperature and Ksp values were plugged into equation (2) to obtain the ΔG value. Plotting a linear graph of equation (3) allowed determining the linear equation from which the ΔH and ΔS terms to be determined algebraically.

**Data:**

*KNO3 mass: 2.051g KNO3 mol: 0.0203*

**Table 1. **Trial Data

| **Trial 1** | **Trial 2** | **Trial 3** | **Trial 4** | **Trial 5** |

**Volume (mL)** | 3.50 | 4.00 | 4.60 | 5.00 | 5.40 |

**[K+], [NO3-]** | 5.79 | 5.07 | 4.41 | 4.06 | 3.75 |

**Temp (K)** | 341.07 | 333.85 | 325.15 | 319.25 | 310.15 |

**Graph 1. ***ln*(Ksp) vs Inverse Temperature

**Results: **

**Table 2. **Thermodynamic results *lit ΔH**value: 34.9 lit ΔS value: 0.20*

| **Trial 1** | **Trial 2** | **Trial 3** | **Trial 4** | **Trial 5** | | **Error** |

**[KNO3]** | 5.79 | 5.07 | 4.41 | 4.06 | 3.75 | | NA |

**Ksp** | 33.59 | 25.72 | 19.45 | 16.46 | 14.11 | | NA |

**ΔG (kJ)** | -9.97 | -9.01 | -8.02 | -7.43 | -6.83 | | NA |

**ΔH (kJ/K)** | NA | NA | NA | NA | NA | 24.92 | 28.59% |

**ΔS (kJ/K)** | NA | NA | NA | NA | NA | 0.108 | 12.24% |

The ΔH and ΔS were 24.92 kJ/K (28.59% error) and 0.108 kJ/K (12.24% error). The Ksp, ΔG, and temperatures are dependent on the trial and are given in Table 2.

**Discussion: **

The experiment was successful in determining the thermodynamic values as intended, with percent errors being below 30%. As expected, the reaction was endothermic with a positive enthalpy value. Also, the relationship between inverse temperature and the ln(Ksp) produced a linear graph that coincided with equation (3) and allowed to calculate the rest of the thermodynamic values.

One legitimate source of error was determining the precise moment of precipitation of solid. Due to their opacity, the flakes were sometimes difficult to spot and were confused with the occasional bubble that was on the edge of the tube. Allowing the temperature to drop lower than the point of equilibrium varied the T value in equation (1) which in turn altered the point of spontaneity of the reaction making it at a lower temperature than actual.

Because thermodynamic values of reactions are simply the negative values of those in the reverse direction, the ΔHo and ΔSo values for the dissociation of KNO3 are 494.6 kJ/mol and -0.133 kJ/molK.

**Questions:**

The data confirmed that the reaction is endothermic and spontaneous only at a certain temperature. As the temperature dropped more salt precipitated. The dissociation of KNO3 results in greater disorder as there are more individual ions. This is confirmed by a positive ΔS value and negative ΔG value.

**Sample Calculations:**

Trial 1.

Ksp = (5.79M)(5.79M) = 33.59

ΔG = -(0.008314 kJ/molK)(341.07K)ln(33.59) = -9.97 kJ/mol

Ln(Ksp) = -2997.5x + 12.244

-ΔH/R = -2997.5

ΔH = 24.92 kJ

ΔS/R = 12.244 kJ

ΔS = 0.108 kJ