Answer these question:
a) the major component in bones is calcium hydroxylapatite, ca5(po3)3oh, with a ksp=7.3*10-34. determine the solubility of bones in pure water.
b) calculate the molar solubility of ca3(po4)2 ( ksp=7.3*10-34 ) in a 0.065 m be3(po4)2 (100% soluble) solution

a) S = 5.9 × 10⁻⁵ M

b) S = 9.2 × 10⁻⁸ M

Explanation:

a) The major component in bones is calcium hydroxylapatite, Ca₅(PO₄)₃OH, with a Ksp=7.3 × 10⁻³⁴. Determine the solubility of bones in pure water.

To determine the solubility (S) of hydroxylapatite in water we will use an ICE chart. We recognize 3 stages: Initial, Change and Equilibrium, and we complete each row with the concentration or change in concentration.

     Ca₅(PO₄)₃OH(s) ⇄ 5 Ca²⁺(aq) + 3 PO₄³⁻(aq) + OH⁻(aq)

I                                         0                   0                  0

C                                      +5S              +3S               +S

E                                        5S                3S                 S

The solubility product Kps is:

Kps = 7.3 × 10⁻³⁴ = [Ca²⁺]⁵.[PO₄³⁻]³.[OH⁻] = (5S)⁵.(3S)³.(S)=84375 . S⁹

S = 5.9 × 10⁻⁵ M

b) Calculate the molar solubility of Ca₃(PO₄)₂ ( Ksp = 7.3 × 10⁻³⁴ ) in a 0.065 M Be₃(PO₄)₂ (100% soluble) solution

First, we have to calculate the concentration of PO₄³⁻ that comes from Be₃(PO₄)₂, which is a strong electrolyte.

        Be₃(PO₄)₂(aq) ⇒ 3 Be²⁺(aq) + 2 PO₄³⁻(aq)

I            0.065 M              0                    0

C         -0.065 M        +3×0.065 M   2×0.065 M

E                0                  0.195 M         0.130 M

Now, 0.130 M will be the initial concentration in the solubilization of Ca₃(PO₄)₂.

          Ca₃(PO₄)₂(s) ⇄ 3 Ca²⁺(aq) + 2 PO₄³⁻(aq)

I                                         0               0.130

C                                     +3S               +2S

E                                       3S            0.130 + 2S

Kps = 7.3 × 10⁻³⁴ = [Ca²⁺]³.[PO₄³⁻]²=(3S)³.(0.130 + 2S)²

In the term (0.130 + 2S)², 2S is very small in comparison with 0.130, so we can neglect it to simplify calculations.

7.3 × 10⁻³⁴ = (3S)³.(2S)² = 108 . S⁵

S = 9.2 × 10⁻⁸ M

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e

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