Question

Some kidney stones form by the precipitation of calcium oxalate monohydrate $\left({\mathrm{CaC}}_2{O}_4\times H_{2}O,K_{\mathrm{sp}}=2.3\times {10}^{-9}\right)$. Theof urine varies from $\mathbf{5}\mathbf{.}\mathbf{5}$ to $\mathbf{7}\mathbf{.}\mathbf{0}$, and the average $\left[{\mathrm{Ca}}^{2+}\right]$in urine is$\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}$.

(a) If the [oxalic acid] in urine is$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{13}}\mathbf{}\mathbf{M}$, will kidney stones form atrole="math" localid="1663267818783" $\mathbf{pH}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{5}$?

(b)at$\mathbf{pH}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{0}$?

(c) Vegetarians have a urineabove$\mathbf{7}$. Are they more or less likely to form kidney stones?

Short answer

1. The Kidney stone will not form at $\mathrm{pH}=5.5$.
2. The Kidney stone will form at $\mathrm{pH}=7.0$.
3. For vegetarian chances of kidney stone formation is more.

Step-by-step solution

Definition of Concept.

Heat Capacity: The amount of heat energy required to raise the temperature of a body by a certain amount is known as heat capacity. The amount of heat in joules required to raise the temperature 1 Kelvin is known as heat capacity (symbol: C) in SI units.

 Explain will the kidney stones form at pH=5.5.

(a)

For precipitation of calcium oxalate $\left({\mathrm{CaC}}_2{\mathrm{O}}_4\right)$(kidney stone) the criteria that must be fulfilled is ${\mathrm{Q}}_{\mathrm{sp}}>{\mathrm{K}}_{\mathrm{sp}}$.

Now,

${\mathrm{Q}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\times \left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]$

Considering the given information:

$\begin{array}{c}\left[{\mathrm{Ca}}^{2+}\right]=2.6\times {10}^{-3}\mathrm{Minurine}\\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\end{array}$

As a result, at pH=5.5,

$\begin{array}{c}\left[{\mathrm{H}}^{+}\right]={10}^{-5.5}\\ =3.16\times {10}^{-6}\end{array}$

Calculating $\left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]$:

role="math" localid="1663268327759" $\begin{array}{rcl}& & {\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\stackrel{\mathrm{K}}{\rightleftarrows }2{\mathrm{H}}^{+}+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\\ & & \begin{array}{c}\left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]=\frac{\mathrm{K}\left[{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}\\ =\frac{\left[3.024\times {10}^{-6}\right]\left[3\times {10}^{-13}\right]}{{\left[3.16\times {10}^{-6}\right]}^2}(K\mathrm{value}\mathrm{given}\mathrm{in}\mathrm{appendix})\\ =9.073\times {10}^{-8}\end{array}\end{array}$

In addition, the ${\mathrm{Q}}_{\mathrm{sp}}$value was calculated as follows:

$\begin{array}{c}{\mathrm{Q}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\times \left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]\\ =\left[2.6\times {10}^{-3}\right]\times \left[9.073\times {10}^{-8}\right]\\ =2.395\times {10}^{-10}\end{array}$

Given that,

${\mathrm{K}}_{\mathrm{sp}}$of calcium oxalate is$2.3\times {10}^{-9}$.

Therefore, ${\mathrm{Q}}_{\mathrm{sp}}<{\mathrm{K}}_{\mathrm{sp}}$and so kidney stone will not form at $\mathrm{pH}=5.5$.

Explain will the kidney stones form at pH=7.0.

(b)

For precipitation of calcium oxalate $\left({\mathrm{CaC}}_2{\mathrm{O}}_4\right)$(kidney stone) the criteria that has to be fulfilled is ${\mathrm{Q}}_{\mathrm{sp}}>{\mathrm{K}}_{\mathrm{sp}}$.

Now,

${\mathrm{Q}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\times \left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]$

Considering the given information:

$\begin{array}{c}\left[{\mathrm{Ca}}^{2+}\right]=2.6\times {10}^{-3}\mathrm{Minurine}\\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\end{array}$

As a result, at pH=7.0

$\begin{array}{c}\left[{\mathrm{H}}^{+}\right]={10}^{-7}\\ =1.0\times {10}^{-7}\end{array}$

Calculate $\left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]$:

data-custom-editor="chemistry" $\begin{array}{rcl}& & {\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\stackrel{\mathrm{K}}{\rightleftarrows }2{\mathrm{H}}^{+}+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\\ & & \begin{array}{c}\left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]=\frac{\mathrm{K}\left[{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}\\ =\frac{\left[3.024\times {10}^{-6}\right]\left[3\times {10}^{-13}\right]}{{\left[1.0\times {10}^{-7}\right]}^2}\\ =9.072\times {10}^{-5}\end{array}\end{array}$

In addition, the${\mathrm{Q}}_{\mathrm{sp}}$value was calculated as,

$\begin{array}{c}{\mathrm{Q}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\times \left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]\\ =\left[2.6\times {10}^{-3}\right]\times \left[9.072\times {10}^{-5}\right]\\ {\mathrm{Q}}_{\mathrm{sp}}=2.4\times {10}^{-7}\end{array}$

Given that,

${\mathrm{K}}_{\mathrm{sp}}$of calcium oxalate is $2.3\times {10}^{-9}$.

Hence, ${\mathrm{Q}}_{\mathrm{sp}}>{\mathrm{K}}_{\mathrm{sp}}$and so kidney stone will form at $\mathrm{pH}=7.0$.

Find the chances for vegetarians have a urine pH above 7 to form kidney stones.

(c)

The reactions are:

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\stackrel{\mathrm{K}}{\rightleftarrows }2{\mathrm{H}}^{+}+{\mathrm{C}}_2{\mathrm{O}}_4^{2-} \\ \begin{array}{c}\left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]=\frac{\mathrm{K}\left[{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\right]}{{\left[{\mathrm{H}}^{+}\right]}^2}\\ {\mathrm{Q}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\times \left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]\end{array} \end{gathered}$$

${\mathrm{Q}}_{\mathrm{sp}}=2.3\times {10}^{-9}\mathrm{atpH}=5.5$and ${\mathrm{Q}}_{\mathrm{sp}}=2.4\times {10}^{-7}$at pH =7.

Hence it has been seen that with the increase in pH the concentration of ${\mathrm{H}}^{+}$decreases and hence $\left[{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right]$increases.

As a result, ${\mathrm{Q}}_{\mathrm{sp}}$will increase and thus easily ${\mathrm{Q}}_{\mathrm{sp}}>{\mathrm{K}}_{\mathrm{sp}}$criteria can be achieved. So precipitation will occur more at higher pH value.

Therefore, the vegetarians are more likely to develop kidney stones.