(a) 400 mg of solid calcium sulfate, CaSO 4 (s), is added to 1.0 L of pure water

Question

(a) 400 mg of solid calcium sulfate, CaSO4 (s), is added to 1.0 L of pure water and the solid begins to dissolve according to the following reaction:

What is the concentration of dissolved sulfate in the water once equilibrium is achieved? If another 400 mg of calcium sulfate is added to the same water, what is the concentration of dissolved sulfate once equilibrium is achieved again? Express the concentrations in mg/L. (Ksp,CaSO4 =2×10-5 at 25°C; Atomic weight: Ca: 40; S:32; O:16)

(b) Calculate the pH and pOH of a 0.001M solution of sodium hydroxide (NaOH) at 25ºC.

Sodium hydroxide is a strong base so assume complete dissociation.

(c) What’s the theoretical oxygen demand for 20 g of propane gas (C3H8)? Determine the volume in liter occupied by the oxygen at 0.21 atm and 30ºC.

(PV=nRT, R: 0.08206 L·atm/(K·mol); atomic weight: C: 12; H:1)

Explanation / Answer

a)

m = 400 mg = 0.4 g V = 1L

mol = masS/MW = 0.4/136.1406 = 0.00293 mol of CaSO4

[CaSO4] = 0.00293/V = 0.00293 M

Note that we have Ksp data

Ksp = [Ca+2][SO4-2]

2*10^-5 = S*S

S = sqrt(2*10^-5 ) = 0.00447 M of CaSO4 will actually be in solution

b)

pH of a 0.001 M NaOH

[NaOH] = [OH-] = 0.001

pOH = -log(OH = -log(0.001)

pOH = 3

pH = 14-3 = 11

c)

find theoretical oxygen demand for

20 g of propane

C3H8 + O2 = CO2 + H2O

balance

C3H8 + 5O2 = 3CO2 + 4H2O

mol of propane = mass/MW =20/44.10 = 0.4535 mol of propane

mol of O2 required -_> 5x0.4535 = 2.2675 mol of O2

mass = mol*MW = 2.2675*32 = 72.56 g of O2

volume

P V= nRT

V= nRT/P = (2.2675)(0.082)(30+273)/(0.21) = 268.277 L O2

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