Question

Calculate the PH and POH of pure water at 25 °C ([H+]=1.00×10-7M).

Answer 1

Since we know that $pH+pOH=pK_w=14$, we proceed to calculate the pH and then the pOH:

$[H^+]=1.00×10^{-7}\,M \\pH=-log[H^+] \\pH=-log(1.00×10^{-7}\,M) \\pH=-\cancel{log(10}^{-7})=-(-7)=7 \\pOH=14-pH=14-7=7$

In conclusion, the pH and pOH of pure water at 25 °C are both equivalent to 7.0

Reference

• Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.