Question

The measured pH values of incoming and outgoing water at a water treatment plant are 7·3 and 8·5 respectively. What is the average pH of water, assuming a linear variation of pH with time?

1. 6·55
2. 7·57
3. 8·12
4. 9·23

Answer 1

Correct Answer - Option 2 : 7·57

Concept: 

pH is a measure of acidity or alkalinity of a solution.

Mathematically, pH is a negative algorithm of the hydrogen ion concentration present in the solution expressed in molarity.

Molarity is defined as the number of moles per liter of solution (mole/liter).

\(pH = \; - {\log _{10}}\left[ {{H^ + }} \right]\)

[H⁺] = 10^-pH

[H⁺] is the concentration of hydrogen ions

Note: Do not use direct pH values as [ H+]

As the variation is linear,

\({\left[ {{{\rm{H}}^ + }} \right]_{{\rm{avg}}}}{\rm{\;}} = \frac{{{{\left[ {{{\rm{H}}^ + }} \right]}_{\rm{i}}} + {{\left[ {{{\rm{H}}^ + }} \right]}_{\rm{o}}}{\rm{\;}}}}{2}\)

Calculations:

Given,

pH of incoming water = 7.3

pH of outgoing water = 8.5

[H⁺]_i of incoming = 10 ^-7.3

[H⁺]_o  of outgoing = 10^-8.5 

\({\left[ {{{\rm{H}}^ + }} \right]_{{\rm{avg}}}}{\rm{\;}} = \frac{{{{\left[ {{{\rm{H}}^ + }} \right]}_{\rm{i}}} + {{\left[ {{{\rm{H}}^ + }} \right]}_{\rm{o}}}{\rm{\;}}}}{2}\)

\({\left[ {{{\rm{H}}^ + }} \right]_{{\rm{avg}}}}{\rm{\;}} = \frac{{{{10}^{ - 7.3}} + {{10}^{ - 8.5}}{\rm{\;}}}}{2}\)

pH = - log[ 2.66× 10^-8 ]

= - log[2.664] + -log[10^-8]

= 7.575

∴ pH_avg = 7.575