Question 20 of 32 Map General Chemistr University Science Books presented by Sap

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Question 20 of 32 Map General Chemistr University Science Books presented by Sapling Lea Donald McQuarrie. Peter A Rock .Ethan Gallogly Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.240 M HCIO(aq) with 0.240 M KOH(aq). The ionization constant for HCIO can be found here. Number (a) before addition of any KOH Number (b) after addition of 25.0 m of KOH Number (c) after addition of 35.0 mL of KoH Number (d) after addition of 50.0 m of KOH Number (e) after addition of 60.0 mL of KOH O Previous Give Up & View Solution Check Answer Next Ext Hint

Explanation / Answer

HClO is a weak acid with dissociation constant of Ka = 2.95x10-8  

HClO ------ > ClO-(aq) + H+(aq) : Ka = 2.95x10-8

(a): Given [HClO], C = 0.240 M

The concentration of H+(aq) can be calculated as

[H+(aq)] = squareroot(Ka x C) = squareroot(2.95x10-8 x 0.240) = 8.41x10-5 M

=> pH = - log[H+(aq)] = - log8.41x10-5 M = 4.08 (answer)

(b): Initial moles of HClO = MxV = 0.240 mol/L x 0.050 L = 0.012 mol

Volume of 0.240 KOH added = 25 mL = 0.025 L

Hence moles of KOH added = MxV = 0.240 mol/L x 0.0250 L = 0.006 mol

Now 0.006 mol KOH will react with 0.006 mol HClO to form 0.006 mol KClO.

After the reaction

moles of HClO remained = 0.012 - 0.006 = 0.006

Hence [HClO] = [KClO] = 0.006 mol / 0.075 L = 0.08 M

Now HClO and KClO will act as buffer slution whose pH can be calculated from Hendersen equation as

pH = pKa + log[KClO] / [HOCl] = 7.53 + log[0.8 / 0.8] = 7.53 (answer)

(c):

Volume of 0.240 KOH added = 35 mL = 0.035 L

Hence moles of KOH added = MxV = 0.240 mol/L x 0.0350 L = 0.0084 mol

Now 0.0084 mol KOH will react with 0.0084 mol HClO to form 0.0084 mol KClO.

After the reaction

moles of HClO remained = 0.012 - 0.0084 = 0.0036

Hence [HClO] = 0.0036 mol / 0.085 L = 0.0424 M

[KClO] = 0.0084 mol / 0.085 L = 0.0988 M

Now HClO and KClO will act as buffer slution whose pH can be calculated from Hendersen equation as

pH = pKa + log[KClO] / [HOCl] = 7.53 + log[0.0988 / 0.0424] = 7.90 (answer)

(d): When 50 mL of KOH is added HOCl is completely neutralized to form 0.012 mol KClO.

=> [KClO] = 0.012 mol / 0.100 L = 0.12 M

Now KClO will undergo hydrolysis and the pH can be calculated from salt hydrolysis equation as

pH = (1/2)x(pKa + pKw + log[KClO]) = (1/2) x (7.53 + 14 + log0.12) = 10.3 (answer)

(d): moles of excess KOH added = 0.240 mol/L x 0.010 L = 0.0024 mol

Total volume of the solution, Vt = 50 mL + 60 mL = 110 mL = 0.110 L

=> [OH-] = 0.0024 mol / 0.110 L = 0.02182 M

=> pOH = -log(0.02182) = 1.66

=> pH = 14 - 1.66 = 12.34 (answer)

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