Window Help Specific x Desmos × w List of b × YG How Max A Buffer x e N Table of

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Window Help Specific x Desmos × w List of b × YG How Max A Buffer x e N Table of x ka hydr eAssignment/takeCovalentActivity.do?locator-assignment-take&takeAssignmentSession; Locator as.. Use the References to access important values If needed for this question. A 32.2 mL sample of a 0.569 M aqueous hypochlorous acid solution is titrated with a 0.333 M aqueous potassium hydroxide solution. What is the pH after 22.8 mL of base have been added? Submit Answer Retry Entire Group 8 more group attempts remaining PreviousNext

Explanation / Answer

Hypochlorous acid is weak acid and its Ka is =3*10-8, pKa= -log Ka= 7.52

The reaction between HOCl and KOH is represented as HOCl+ KOH ------>KOCl+ H2O

Theoretical molar ratio of HOCl and KOH= 1:1

moles= Molarity* Volume in liters, 1000ml =1 L,

moles of HOCl= 0.569*32.2/1000= 0.0183, moles of KOH in 22.8 ml of 0.333M= 0.333*22.8/1000 =0.0076

Actual molar ratio of HOCl : KOH= 0.0183:0.0076 =0.0183/0.0076 :0.0076/0.0076= 2.41:1

So excess is HOCl and all the KOH reacts and forms KOCl whch is conjugate of acid HCOL which gives OCl- ions

hence moles of KOCl formed= 0.0076, moles of HOCl remaining = 0.0183-0.0076=0.011

Volume of solution after mixing the reactants= 32.2+22.8= 55ml= 55/1000L

Concentrations : KOCl=0.0076/(55/1000), HOCl=0.011/(55/1000)

From Henderson-Hasselbalch equation, pH= pKa+ log{ [Conjugate base]/[Acid]}

pH= 7.52+ log (0.0076/0.011) =7.36

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