Q159CP - Chemistry: Molecular Nature Of Matter And Change

Question

Human urine has a normal pH of 6.2. If a person eliminates an average of 1250 mL of urine per day, how many ${\mathbf{\text{H}}}^{\mathbf{\text{ + }}}$  ions are eliminated per week?

Step-by-Step Solution

Answer

 $3.3\times {10}^{18}{\text{H}}^{\text{ + }}\text{ ions}$ are eliminated

Step 1: To find how many H  +  ions are eliminated

First solve for the $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]$ from the given pH

$\begin{array}{rcl}& & \text{ pH = } \text{ - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\\ & & \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = } {\text{10}}^{\text{ - pH}}\\ & & {\text{ = 10}}^{\text{ - 6.2}}\\ & =& 6.31\times {10}^{-7}\mathrm{M}\end{array}$

Solve for the number of ions by converting it to moles and using Avogadro's number.

$$\begin{gathered} {\text{no. of H}}^{\text{ + }}\text{ions = } 6.32\times {10}^{-7}\mathrm{M}\times 1.250 \mathrm{L}\times \frac{6.022\times {10}^{23} {\displaystyle {{\displaystyle \text{H}}}^{\text{ + }}}{\displaystyle \text{ions }}}{1 \mathrm{mol} \mathrm{H}} \\ {\text{ no. of H}}^{\text{ + }}\text{ions = } 4.75\times {10}^{17}{\mathrm{H}}^{\text{ + }}\mathrm{ions} \end{gathered}$$

$4.75\times {10}^{17} {\mathrm{H}}^{+}$ ions are eliminated per day. Multiply that by 7 days a week $3.3\times {10}^{18}{\mathrm{H}}^{+}$.  ions are eliminated from an average person's body per week.

Hence,  $3.3\times {10}^{18}{\mathrm{H}}^{+}$are eliminated.