a.You have 40.0 mL of a 0.100 M solution of HClO, with a Ka of 3.0...

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a.You have 40.0 mL of a 0.100 M solution of HClO, with aK_a of 3.0 x 10^-8. b. What is the pH of thesolution after the addition of 30.0 mL of 0.200 M NaOH?

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4.0 Ratings (542 Votes)

millimoles of HClO = 40 x 0.1 = 4

millimoles of NaOH = 30 x 0.2 = 6

HClOÂ Â + NaOH ----------------------> NaClO + H2O

4Â Â Â Â Â Â Â Â Â Â Â Â Â Â 6Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â 0 --------------------> initial

0Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 4Â Â Â Â Â Â Â Â Â Â Â Â 4 ------------------> after reaction

strong base NaOH remained in the solution .

so [OH] = millimoles / total volume

Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 / (40 +30)

Â Â Â Â Â Â Â Â Â Â Â Â = 0.0286 M

pOH = -log [OH-] = -log (0.0286)

pOH = 1.54

pH + pOH = 14

pH = 12.46

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