Question 13 of 36 University Science Books presented by Sapling Learning Donald

Question

Question 13 of 36 University Science Books presented by Sapling Learning Donald McQuarrie Peter A. Rock-Ethan Gallogly Calculate the pH for each of the foll with 0.180 M KOH(aq). The ionization constant for HCIO can be found here. owing cases in the titration of 50.0 mL of 0.180 M HCIO(aq) Number (a) before addition of any KOH Number (b) after addition of 25.0 mL of kOH Number (o) after adion of 35.0 m of KOH Number (d) after addition of 50.0 mL of KOH Number (e) after addition of 60.0 mL of KOH

Explanation / Answer

pH = 1/2 (pKa - log C)

pH = 1/2 ( -log(4 * 10^-8) - log (50 * 0.18/1000))

pH = 4.722

after addition 25 mLof KOH,

pH = 7 + 1/2 (pKa +log C)

pH = 7 + 1/2 ( -log(4*10^-8) + log ((25*0.18)/(25+50))

pH = 10.088

after addition of 35 mL KOH,

pH = 7 + 1/2 (pKa +log C)

pH = 7 + 1/2 ( -log(4*10^-8) + log ((35*0.18)/(35+50))

pH = 10.134

after addition of 50 mL KOH,

pH = 7 + 1/2 (pKa +log C)

pH = 7 + 1/2 ( -log(4*10^-8) + log (50*0.18/100))

pH = 10.176

after addition of 60 mL KOH,

Number of moles of base remaining = ((60*0.18/1000) - (50*0.18/1000)))

                                                                   = 0.0018

pOH = -log(0.0018) = 2.744

pH = 14 - 2.744 = 11.256

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