2. Kidney stones are pieces of Calcium Oxalate (CaC204) that grow in the kidneys
ID: 548946 • Letter: 2
Question
2. Kidney stones are pieces of Calcium Oxalate (CaC204) that grow in the kidneys and are excreted through urine. The Kb of C2042- is 1.639 × 10-10 and the Ksp of CaC204 is 3.0 × 10-9. (a) Write down the solubility equilibrium for CaC204 (b) Write down the hydrolysis equilibrium for C2042 (c) If the normal pH of kidney fluid is 8.2, predict whether the formation of kidney stones can be minimized by increasing or decreasing the pH of the fluid present in the kidneys (d) Why does it make sense that people who ingest Calcium supplements have a higher risk of kidney stone formation?Explanation / Answer
a) CaC2O4(s) <---------> Ca2+(aq) + C2O42-(aq)
Ksp = [Ca2+] [ C2O42-] = 3.0×10^-9
b) C2O42-(aq) + H2O(l) <--------> HC2O4-(aq) + OH-(aq)
Kb = [ HC2O4-] [OH-]/[C2O42-] = 1.639×10^-10
c) Combining the above two equation
CaC2O4(s) + C2O42-(aq) + H2O(l) <--------> Ca2+(aq) + C2O42- (aq) + HC2O42-(aq) + OH-(aq)
K =( [Ca2+][C2O42-]/[CaC2O4] )([HC2O4-][OH-]/[C2O42-]
= Ksp × Kb
= 3.0×10^-9 × 1.639 × 10^-10
= 4.92 × 10^-19
Net ionic equation of the combined equation is
CaC2O4(s) + H2O(l) <-----> Ca2+ (aq) + HC2O4-(aq) + OH-(aq)
K = [Ca2+] [ HCO4- ] [ OH-] = 4.92 × 10^-19
This the condition for saturation of CaC2O4 in H2O
pH = 8.2
pOH = 14 - 8.2 = 5.8
[ OH- ] = 1.58×10^-6M
[ OH- ] = [ HC2O4- ]
Therefore,
[ Ca2+] = 4.92 ×10^-19/(1.58×10^-6)^2
= 1.97 ×10^-7M
Suppose , decreasing the pH , say pH = 7.5
pOH = 6.5
[ OH- ] = 3.16×10^-7
Therefore,
[ Ca2+ ] = 4.92 × 10^-19/(3.16 ×10^-7)^2
= 4.93× 10^-6M
[ Ca2+ ] concentration increasing i.e CaC2O4(s) formation decreasing
Therefore,
Formation of kidney stone minimized by decreasing pH of flouid present in kidneys
d) K = [ Ca2+ ] [ HC2O4-] [ OH-] = 4.92 ×10^-19
Therefore,
when pH is fixed
increasing [ Ca2+ ] will increase the formation of CaC2O4(s) i.e Kidney stone
Therefore, peple who ingset calcium suplements have higher risk of kidney stone formation
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