Question

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

Short answer

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

Step-by-step solution

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

role="math" localid="1663313092236" $\begin{array}{rcl}& & \text{pH =}\text{- log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ & & \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{- pH}}\\ & & {\text{= 10}}^{\text{- 6.2}}\\ & =& 6.31\times {10}^{-7}\mathrm{M}\end{array}$

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

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

$4.75\times {10}^{17}{\mathrm{H}}^{+}$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.