Calcium sulfate is only sparingly soluble. caso4(s) ⇌ ca2+(aq) + so42-(aq) for this type of dissolution reaction the equilibrium constant, also known as the solubility product, is denoted ks. in the reaction above, ks = 3.9 x 10-6. when an excess of the solid is dissolved in water what is the maximum concentration of ca2+(aq) in mol l-1?

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as

We are given:

Solubility product of

By stoichiometry of the reaction:

1 mole of gives 1 mole of and 1 mole of .

When the solubility of is S moles/liter, then the solubility of will be S moles\liter and solubility of will be S moles/liter.

Expression for the equilibrium constant of will be:

Hence, the maximum concentration of  is

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j