# Chemical Reaction and Solution

Topics: Chemical reaction, Chemistry, Acid Pages: 12 (2464 words) Published: February 18, 2013
AP Chem Exam - ‘98

1.Solve the following problem related to the solubility equilibria of some metal hydroxides in aqueous solution.

(a)The solubility of Cu(OH)2(s) is 1.72 x10–6 g/100. mL of solution at 25° C.

(i)Write the balanced chemical equation for the dissociation of Cu(OH)2(s) in aqueous solution.

Cu(OH)2 Cu 2+ + 2 OH –

(ii)Calculate the solubility (in mol/L) of Cu(OH)2 at 25 °C.

(1.72 x10–6 g/0.100 L)(1 mol/97.5 g) = 1.76 x10–7 mol/L

(iii)Calculate the value of the solubility-product constant, Ksp, for Cu(OH)2 at 25 °C.

Ksp = [Cu 2+][OH –]2 = [1.76 x10–7][3.53 x10–7]2 = 2.20 x10–20

(b)The value of the solubility-product constant, Ksp, for Zn(OH)2 is 7.7 x10–17 at 25 °C.

(i)Calculate the solubility (in moles per liter) of Zn(OH)2 at 25 °C in a solution with a pH of 9.35.

Zn(OH)2 Zn 2+ + 2 OH –

Ksp = [Zn 2+][OH –]2
pH 9.35 = pOH 4.65; [OH –] = 10–4.65 = 2.24 x10–5 M

Zn2+ = Ksp/[OH -]2 = 7.7x10-17/[2.24 x10-5]2 = 1.5 x10-7 M

(ii)At 25˚C, 50.0 mL of 0.100-M Zn(NO3)2 is mixed with 50.0 mL of 0.300-M NaOH. Calculate the molar concentration of Zn2+ (aq) in the resulting solution once equilibrium has been established. Assume that volumes are additive.

[Zn2+]init = (0.100M )(0.050 L) = 0.0500 M
[OH–]init = (0.300M)(0.050 L) = 0.150 M

X = conc. loss to get to equilibrium

Ksp = 7.7 x10–17 = (0.0500 – X)(0.150 – 2X)2
[Zn 2+] = 0.0500 – X = 3.1 x10–14M

2.An unknown compound contains only the three elements C, H, and O. A pure sample of the compound is analyzed and found to be 65.60% C and 9.44% H by mass.

(a)Determine the empirical formula of the compound.

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(C3.5H6O1)2 = C7H12O2

(b)A solution of 1.570 g of the compound in 16.08 g of camphor is observed to freeze at a temperature 15.2° C below the normal freezing point of pure camphor. Determine the molar mass and apparent molecular formula of the compound. (The molal freezing-point depression constant, Kƒ, for camphor is 40.0 kg-K-mol–1.)

MM = [1.570 g)(40 kg-° C/mol)]/[(0.01608 kg)(15.2° C)] = 257 g/mol

C7H12O2 = 128 g/mol 257/128 = 2 (C7H12O2)2 = C14H24O4

(c)When 1.570 grams of the compound is vaporized at 300˚C and 1.00 atm, the gas occupies a volume of 577 milliliters. What is the molar mass of the compound based on this result?

MM = (1.570 g)(0.0821 L-atm/mol-K)(573 K)/(1.00 atm)(0.577 L) = 128 g-mol-1

(d)Briefly describe what occurs in solution that accounts for the difference between the results obtained in parts (b) and (c).

since the apparent molar mass from freezing point change is twice that determined by the vapor, the molecules of the compound must associate in camphor to form dimers.

3.C6H5OH(s) + 7 O2(g) 6 CO2(g) + 3 H2O(l)

When a 2.000-g sample of pure phenol, C6H5OH(s) is completely burned according to the equation above, 64.98 kJ of heat is released. Use the information in the table below to answer the questions that follow.

|Substance |∆H˚ƒ; at 25 °C (kJ/mol) |S˚, at 25 °C (J/mol-K) | |C(graphite) |0.00 |5.69 | |CO2(g) |–393.5 |213.6 | |H2(g) |0.00 |130.6 | |H2O(l) |–285.85 |69.91 | |O2(g) |0.00 |205.0 | |C6H5OH(s) |? |144.0...

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