Trial Number Original mass of sample (g) Volume of water (mL) added to rinse sal
ID: 1020060 • Letter: T
Question
Trial Number
Original mass of sample (g)
Volume of water (mL) added to rinse salt
Number of drops of water added to saturation
Total volume
(mL) water added
Moles of salt added
Molarity
(M) of salt solution
NaCl – 1
2.5
2
72
2.2
0.043
20
NaCl – 2
2.5
2
72
2.2
0.043
20
NaCl – 3
2.5
2
74
2.3
0.043
19
KCl – 4
2.5
2
52
1.6
0.034
21
KCl – 5
2.5
2
54
1.7
0.034
20
KCl – 6
2.5
2
52
1.6
0.034
21
1.Convert concentration of salt at the end of each of the six trials into molarities of KCl and NaCl,
2.Write the dissociation reactions for both salts.
3.Determine the molarity of ions in the solution.
4.Write the formula for the Ksp of each salt.
5.Calculate the Ksp for NaCl and Kcl.
6.The theoretical value for the Ksp of KCl at 0°C is 12, and, at the same temperature, the Ksp for NaCl is 32. Determine the percent error between these theoretical values and your experimental values?
7.Why do you think the instructions for this lab included using chilled distilled water? In other words, why didn't you simply use room temperature water?
8.Predict how the Ksp values of both salts would change if this experiment were done at temperatures above 90°C. Explain how the values would change when comparing both salts to each other.
Trial Number
Original mass of sample (g)
Volume of water (mL) added to rinse salt
Number of drops of water added to saturation
Total volume
(mL) water added
Moles of salt added
Molarity
(M) of salt solution
NaCl – 1
2.5
2
72
2.2
0.043
20
NaCl – 2
2.5
2
72
2.2
0.043
20
NaCl – 3
2.5
2
74
2.3
0.043
19
KCl – 4
2.5
2
52
1.6
0.034
21
KCl – 5
2.5
2
54
1.7
0.034
20
KCl – 6
2.5
2
52
1.6
0.034
21
Explanation / Answer
1) Let me show you how we calculate the molarity of NaCl with trial 1.
Molarity of NaCl = moles of NaCl/volume of solution in L = moles of NaCl/total volume of water added in L = 0.043 mole/[(2.2 mL)*(1 L/1000 mL)] = 19.545 mole/L 20.0 M
Average molarity of NaCl = [(20 + 20 + 19) M/3] = 19.67 M
Average molarity of KCl = [(21 + 20 + 21) M/3] = 20.67 M (ans)
2) The dissociation reactions are
NaCl (aq) -----------> Na+ (aq) + Cl- (aq)
KCl (aq) -----------> K+ (aq) + Cl- (aq)
I have used a single directional arrow since both these salts are strong electrolytes in aqueous solution and dissociate completely into ions.
3) From the two equations above, we can see that a 1:1 molar equation exists between the electrolyte and the dissociated ions. Therefore, for NaCl,
NaCl (aq) --------> Na+ (aq) + Cl- (aq)
[Na+] = 19.67 M
and [Cl-] = 19.67 M
Similarly, for KCl, we have [K+] = 20.67 M and [Cl-] = 20.67 M (ans)
4) We usually calculate Ksp for sparingly soluble salts; however NaCl and KCl are both extremely soluble. The expressions for Ksp are
Ksp (NaCl) = [Na+][Cl-] …….(1)
Ksp (KCl) = [K+][Cl-] ……(2)
5) Ksp (NaCl) = (19.67).(19.67) = 386.9098 = 3.869*102
Ksp (KCl) = (20.67).(20.67) = 427.2489 = 4.272*102 (ans)
6) The theoretical values of Ksp for NaCl and KCl are 12 and 32 at 0°C.
Therefore,
% error (NaCl) = [(experimental value – theoretical value)/theoretical value]*100 = [(386.9098 – 12)/12]*100 = 3124.248
% error (KCl) = [(427.2489 – 32)/32]*100 = 1235.153 (ans)
Such a large error creeps in because we are trying to calculate the Ksp for a highly soluble electrolyte. Ksp is the ratio of the dissociated and undissociated fractions of a sparingly soluble electrolyte. For a highly soluble electrolyte, there is no undissociated fraction and hence the said quantity is actually unity. Moreover, Ksp, being an equilibrium constant, is highly sensitive to temperature changes. The theoretical values are given at 0°C and we are performing the experiment possibly at room temperature which is around 22-25°C. Thus, the high error percentage.
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