6.50×10^(-4) M HOCl 7.50×10^(-4) M NaOCl buffer pH

Calculate the final, equilibrium pH of a buffer

that initially contains 6.50×10^(-4) M HOCl and 7.50×10^(-4) M NaOCl.

The Ka of HOCl is 3.0×10^(-5).

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Henderson-Hasselbalch 식

pH = pKa + log([짝염기]/[약산])

( 참고 https://ywpop.tistory.com/1926 )

= –log(3.0×10^(-5)) + log([7.50×10^(-4)] / [6.50×10^(-4)])

= 4.58503

답: pH = 4.59

[키워드] HOCl buffer 기준, HOCl 완충용액의 pH 기준