Question

The pH of normal human urine is in the range of 5.5 to 6.5. Compute the range of H₃O⁺ concentration and the range of the OH^- concentration in normal urine.

Short answer

The range ofion will be$\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{<}\left[H_3{O}^{+}\right]\mathbf{<}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}$ , and the range of $O{H}^{-}$ion will be$\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{<}\left[O{H}^{-}\right]\mathbf{<}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}$

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Step-by-step solution

Understanding the concept of pH of an acid-base.  

The negative logarithm of ${\mathit{H}}^{\mathbf{+}}$ ion concentration is used to calculate pH. It's a scale that is used to measure acids and bases.

$\mathit{p}\mathit{H}\mathbf{=}\mathbf{-}\mathit{l}\mathit{o}\mathit{g}\left[H_3{O}^{+}\right]$

Also we know that $\mathit{p}\mathit{H}\mathbf{+}\mathit{p}\mathit{O}\mathit{H}\mathbf{=}\mathbf{14}$

Applying the concept and calculation of concentration of hydronium ion

Given range of pH is 5.5 to 6.5.

Now to calculate the range of concentration of H₃O⁺ ion, first calculate the concentration at the pH value of 5.5

$$\begin{gathered} pH=-\log \left[H_3{O}^{+}\right] \\ 5.5=-\log \left[H_3{O}^{+}\right] \\ \left[H_3{O}^{+}\right]={10}^{-5.5} \\ \left[H_3{O}^{+}\right]=3\times {10}^{-6} \end{gathered}$$

Now to calculate the range of concentration of H₃O⁺ ion, first calculate the concentration at the pH value of 6.5

$$\begin{gathered} pH=-\log \left[H_3{O}^{+}\right] \\ 6.5=-\log \left[H_3{O}^{+}\right] \\ \left[H_3{O}^{+}\right]=3\times {10}^{-7} \end{gathered}$$

Applying the concept and calculation of concentration of hydroxide ion

Given range of pOH is 5.5 to 6.5.

Now to calculate the range of concentration of$O{H}^{-}$ion, first, calculate the concentration at the pOH value of 5.5

$$\begin{gathered} \left[H_3{O}^{+}\right]=3\times {10}^{-6} \\ \left[O{H}^{-}\right]=\frac{1\times {10}^{-14}}{3\times {10}^{-6}} \\ \left[O{H}^{-}\right]=3\times {10}^{-9} \end{gathered}$$

Now to calculate the range of concentration of OH^- ion, first, calculate the concentration at the pOH value of 6.5

$$\begin{gathered} \left[H_3{O}^{+}\right]=3\times {10}^{-7} \\ \left[O{H}^{-}\right]=\frac{1.0\times {10}^{-14}}{3\times {10}^{-7}} \\ \left[O{H}^{-}\right]=3\times {10}^{-8} \end{gathered}$$