Question

Hydroxyapatite, Ca 10 ( PO 4 ) 6 ( OH ) 2 , has a solubility constant of Ksp = 2.34 × 10 − 59 , and dissociates according to Ca 10 ( PO 4 ) 6 ( OH ) 2 ( s ) − ⇀ ↽ − 10 Ca 2 + ( aq ) + 6 PO 3 − 4 ( aq ) + 2 OH − ( aq ) Solid hydroxyapatite is dissolved in water to form a saturated solution. What is the concentration of Ca 2 + in this solution if [ OH − ] is fixed at 3.90 × 10 − 6 M ?

Answer 1

Answer:

Hydroxyapatite, Ca10(PO4)6(OH)2 , has a solubility constant of Ksp = 2.34×10−59 , and dissociates according to Ca10(PO4)6(OH)2(s)↽−−⇀10Ca2+(aq)+6PO3−4(aq)+2OH−(aq) Solid hydroxyapatite is dissolved in water to form a saturated solution.

Explanation:

Answer 2

Answer: The concentration of $Ca^{2+}$ ions in the solution is $7.81\times 10^{-4}M$

Explanation:

We are given:

Concentration of hydroxide ion = $3.90\times 10^{-6}M$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The equation for the ionization of the hydroxyapatite is given as:

  $Ca_{10}(PO_4)_6(OH)_2(s)\leftrightharpoons 10Ca^{2+}(aq.)+6PO_4^{3-}(aq.)+2OH^-(aq.)$

                                            10s                  6s         $(3.90\times 10^{-6}+2s)$

The expression for the solubility product of hydroxyapatite will be:

$K_{sp}=[Ca^{2+}]^10[PO_4^{3-}]^6[OH^-]^2\\\\K_{sp}=(10s)^{10}\times (6s)^6\times (3.90\times 10^{-6}+2s)^2=4.6656\times 10^{14}\times s^{16}\times (3.90\times 10^{-6}+2s)^2$

We are given:

$K_{sp}=2.34\times 10^{-59}$  

Putting values in above equation, we get:  

$2.34\times 10^{-59}=4.6656\times 10^{14}\times s^{16}\times (3.90\times 10^{-6}+2s)^2\\\\s=7.81\times 10^{-5},-7.86\times 10^{-5}$

Neglecting the negative value of 'x' because concentration cannot be negative.

So, the concentration of calcium ions in the solution = 10s = $[10\times (7.81\times 10^{-5})]=7.81\times 10^{-4}M$

Hence, the concentration of $Ca^{2+}$ ions in the solution is $7.81\times 10^{-4}M$