Answer to Question #235938 in General Chemistry for Smash

Question #235938

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


1
Expert's answer
2021-09-12T00:42:35-0400

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


[H+]=1.00×107MpH=log[H+]pH=log(1.00×107M)pH=log(107)=(7)=7pOH=14pH=147=7[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.

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