Chapter 18.3, Problem 23PP

Interpretation Introduction

Interpretation:

The number of $H^{+}$ and $O{H}^{-}$ ions are to be calculated in given volume.

Concept introduction:

Concept of ionic product is used.

At 298 K $\left[H^{+}\right]=\left[O{H}^{-}\right]=1\times {10}^{-7}M$

This information can be used to calculate concentration at different volume.

Answer to Problem 23PP

Number of $H^{+}$ ions = Number of $O{H}^{-}$ ions = $1.8\times {10}^{16}ions$

Explanation of Solution

At 298 K $\left[H^{+}\right]=\left[O{H}^{-}\right]=1\times {10}^{-7}M$

This implies ${10}^{-7}mol$ are present in 1 litre water.

Therefore, $300mLwaterwillcontain=\frac{300mL}{1000mL}\times {10}^{-7}$

= $3\times {10}^{-8}mol$

These moles can be converted into number of ions as:

Number of $H^{+}$ ions = Number of $O{H}^{-}$ ions = $3\times {10}^{-8}\times 6.023\times {10}^{23}$

= $1.8\times {10}^{16}ions$

Conclusion

Thus,

Number of $H^{+}$ ions = Number of $O{H}^{-}$ ions = $1.8\times {10}^{16}ions$