Question

1) Hydroxylapatite, Ca10(PO4)6(OH)2, has a solubility constant of Ksp = 2.34 × 10-59, and dissociates according to Ca10(PO4)6OH2 <======> 10Ca(2+) + 6PO4(3-) + 2OH-

Solid hydroxylapatite is dissolved in water to form a saturated solution. What is the concentration of Ca2 in this solution if [OH–] is somehow fixed at 1.20 × 10-6 M?

2) Determine the concentration of an HBr solution if a 40.00 mL aliquot of the solution yields 0.6828 g AgBr when added to a solution with an excess of Ag ions. The Ksp of AgBr is 5.0 × 10–13.

Answer 1

Answer 2

des given 9927272721 III Contiontration ok Ön is fixed at egelb?? 46.9 X 10 CM Ksp= 2.34 X10-59 solution dissolved eq. Caro (POW), (OH), (S) = 10ca*caq) + 6 Pou? Tag) + 2 oñcay) at saturation los 65 409X106 so ksp = [Ca2+70 [Pows] [on]² 4910 2.34X1059 = (103) (65) 4.9x10-6 (6 8 10" X 4.986063/6 = 2.34X10-59 gl6 = 2.3481059 (6) ² x 100 x 4.9X106 S = 1024 x 10-69 5o = 0.0001024 x 10-64 s = 0. 563 810-4 e11111111111 [Ca 2+] = 10s= 5063 810-4 M Thank you, May it help you

Answer 3

Cao (PO 4) 6 (OH), (s) — 10 cart laq) + 6 P04 + 2011 (6) Concentrations [n] [0.64] [6.2x10-6] then solubility constant = [Ca 2+ ] 1° [ P043-] 4 [on]² * 2.34x10-59 = [u] 0 [0.62] * [6.2 x 16-632 t n 16 = 2.34X10-59 6.2 X 6.2 % 10-12 (0.6) 6 - n = 10304743 810-47 D na 1.174810-3 M. So, [cat] = [1.174810-3 M

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Answer 4

Answer:-

The answer is given in the image,

C_10 (PO_4)_6 (OH)_2, ksp = 2.34 times 10^-59 C_10 (PO_4)_6 (OH)_2 rightharpoon overleft harpoon 10 Ca^2 + 6 PO_4^3- + 2 OH^- 10s 6s 2s so, [OH^-] = 5.8 times 10^-6 M k_sp = [Ca^2+]^10 [PO_4 ^3-]^6 [OH^-]^2 2.34 times 10^-59 = (105)^10 (6s)^6 (5.8 times 10^-6)^2 2.34 times 10^-59 = 4.6656 times 10^14 s^16 times (5.8 times 10^-6)^2 2.34 times 10^-59 = 1.57 times 10^4 s^16 s^16 = 1.49 times 10^-63 s = 1.18 times 10^-4 M so, [Ca^2+] = 10 s [Ca^2+] = 10 times 1.18 times 10^-4 [Ca^2+] = 1.18 times 10^-3 M