1. When you did the 9 Bottles experiment in General Chemistry 1, explain why you
ID: 1039645 • Letter: 1
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1. When you did the 9 Bottles experiment in General Chemistry 1, explain why you did not observe a precipitate when you mixed equal volumes of 0.01 M BaCl2 is mixed with 0.1 M Pb(NOs)2. The Ksp for PbCl2 is 2.4 x 104. 2. Barbituric acid, HC4H3N203, a monoprotic acid, is used to prepare various barbiturate drugs. (a) Calculate the concentration of hydrogen ion and pH in 100.0 ml of 0.250 M solution of the acid. The value of Ka is 9.8 x 10-5. (b) If 25.0 ml of 0.50 M NaOH is added to the above solution, what is the pH of the resulting solution? (c) After a total of 50.0 ml of 0.50 M NaOH is added, what is the pH of this solution now? (d) What is the pH of the resulting solution after 55.0 ml of 0.50 M NaOH? 3. What is solubility? What is the solubility product? How will intermolecular forces (IMF) affect the solubility of KHTar? How will the effect of IMF be verified in next week's experiment?Explanation / Answer
1)
All Nitrate (NO3-) salts are soluble.
Barium chloride(BaCl2) is soluble, but lead chloride (PbCl2) is insoluble.
So, the possible precipitate is PbCl2.
If the reaction quotient (Q) of PbCl2 is greater than solubility product (Ksp) of PbCl2, precipitate of PbCl2 will form.
Since the concentration of BaCl2 is 0.01 M, the concentration of Cl- is 2x(0.01 M) = 0.02 M.
Since the concentration of Pb(NO3)2 is 0.1 M, the concentration of Pb2+ is 0.1 M.
Since the volume of two solutions are equal,the final concentrations are;
[Cl-] = 0.02 M/2 = 0.01 M
[Pb2+] = 0.1 M/2 = 0.05 M
Q for PbCl2 is,
Q = [Pb2+][Cl-]2 = (0.05)x(0.01)2 = 5x10-6
The Q value is less than Ksp value (2.4x10-4).
So, no precipitate will form.
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