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Chapter 15: Q49E (page 874)

Question: About 50% of urinary calculi (kidney stones) consist of calcium phosphate,\(C{a_3}{\left( {P{O_4}} \right)_2}\). The normal mid-range calcium content excreted in the urine is 0.10 g of \(C{a^{2 + }}\) per day. The normal mid-range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form?

Short Answer

Expert verified

The maximum concentration of phosphate ion \(\left( {PO_4^{3 - }} \right) = 4 \times {10^{ - 9}}\).

Step by step solution

01

Calculate maximum concentration:

\({\rm{C}}{{\rm{a}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2}(s) \leftrightarrow 3{\rm{C}}{{\rm{a}}^{2 + }}(aq) + 2{\rm{P}}{{\rm{O}}_4} \sim 3 - (aq)\)

Concentration of calcium in urine:

\(\begin{array}{l}\left( {{\rm{C}}{{\rm{a}}^{2 + }}} \right) = \frac{{m \times M}}{V}\\ = \frac{{0.1{\rm{g}} \times 0.025{\rm{mol}}/{\rm{g}}}}{{1.4{\rm{L}}}}\\ = 1.8 \times {10^{ - 3}}{\rm{M}}\end{array}\)

\(\begin{array}{l}{K_{sp}} = {\left( {C{a^{2 + }}} \right)^3}{\left( {PO_4^{3 - }} \right)^2}\\ = 1 \times {10^{ - 25}}\end{array}\)

\(\begin{array}{l}\left( {PO_4^{3 - }} \right) = \sqrt {\frac{{1 \times {{10}^{ - 25}}}}{{{{\left( {1.8 \times {{10}^{ - 3}}} \right)}^3}}}} \\ = 4 \times {10^{ - 9}}\end{array}\)

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