Calculate the pH of a colution that is 0.41M in HOCL and 0.50M in NaOCL. Ka(HOCL)=3.2X10^-8?
Answer:
HOCL<==>ClO- + H+
pH=pKa+[ClO-]/[HOCl]
[ClO-]approximately=0.5M
[HOCl]approximately=0.41M
pH=7.49+log(0.5/0.41)
pH=7.49+0.086
pH=7.58
pH = pK + log( [NaOCl]/[HOCl]
pK = log 1/Ka =7.49
pH = 7.49 +log 0.5/0.41 =0.086+7.49= 7.58
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