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Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq). (a) before addition of any KOH (b) after addition of 25.0 mL of KOH (c) after addition of 35.0 mL of KOH (d) after addition of 50.0 mL of KOH (e) after addition of 60.0 mL of KOH

Deborah Edwards

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Ross Pratt on April 7, 2018

(a) before addition of any KOH : when we use the Ka equation & Ka = 4 x 10^-8 : Ka = [H+]^2 / [ HCIO]by substitution:4 x 10^-8 = [H+]^2 / 0.21[H+]^2 = (4 x 10^-8) * 0.21 = 8.4 x 10^-9[H+] = √(8.4 x 10^-9) = 9.2 x 10^-5 Mwhen PH = -㏒[H+] PH = -㏒(9.2 x 10^-5) = 4 (b)After the addition of 25 mL of KOH: this produces a buffer solution, So that, we will use the Henderson-Hasselbalch equation for the PH:PH = Pka +㏒[Salt]/[acid]in the first place, we have to get the moles of HCIO= molarity * volume =0.21 M * 0.05 L = 0.0105 molesthen, moles of KOH = molarity * volume = 0.21 * 0.025 =0.00525 mol ∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525, and when the total volume = 0.05 L + 0.025 L = 0.075 LSo molarity of HCIO = moles HCIO remaining / total volume = 0.00525 / 0.075 =0.07 Mand molarity of KCIO = moles KCIO / total volume = 0.00525 / 0.075 = 0.07 Mand when Ka = 4 x 10^-8 ∴Pka =-㏒Ka = -㏒(4 x 10^-8) = 7.4 for replacement in the H-H equation:PH = 7.4 + ㏒(0.07/0.07)∴PH = 7.4 (c) after the addition of 35 mL of KOH:we're going to use the H-H equation again as we have a buffer solution of PH = Pka + ㏒[salt/acid]in the first place, we have to get the moles HCIO = molarity * volume = 0.21 M * 0.05 L = 0.0105 molesthen moles of KOH = molarity * volume = 0.22 M* 0.035 L =0.0077 mol ∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5when the total volume = 0.05 L + 0.035 L = 0.085 L∴ the molarity of HCIO = moles HCIO remaining / total volume = 8 x 10^-5 / 0.085 = 9.4 x 10^-4 Mand the molarity of KCIO = moles KCIO / total volume = 0.0077 M / 0.085 L = 0.09 mby'a substitution:PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)∴PH = 8.38 (D)After the addition of 50 mL:from the solutions above, you can see that 0.0105 mol HCIO react with 0.0105 mol of KOH to produce 0.0105 mol KCIO that dissolve in 0.1 L (0.5 L+0.5 L) of the solution.the molarity of KCIO = moles KCIO / total volume = 0.0105 mol / 0.1 L = 0.105 Mwhen Ka = KW / Kb∴Kb = 1 x 10^-14 / 4 x 10^-8 = 2.5 x 10^-7by using the Kb expression:Kb = [CIO-] [OH-] / [KCIO]when [CIO, -] =[OH-] so that we can substitute for [OH-] in place of [CIO-]Kb = [OH-]^2 / [KCIO] 2.5 x 10^-7 = [OH-]^2 /0.105∴[OH-] = 0.00016 MPOH = -㏒[OH-]∴POH = -㏒0.00016 = 3.8∴PH = 14 - POH =14 - PH 3.8 = 10.2 (e) after the addition of 60 mL of KOH:when KOH neutralized all the HCIO so, to get the molarity of KOH solutionM1*V1= M2*V2 where M1 is the molarity of KOH solutionV1 is the total volume = 0.05 + 0.06 = 0.11 LM2 = 0.21 M V2 is the excess added volume of KOH = 0.01 Lso by substitution:M1 * 0.11 L = 0.21*0.01 L∴M1 =0.02 M∴[KOH] = [OH-] = 0.02 M∴POH = -㏒[OH-] = -㏒0.02 = 1.7∴PH = 14 - POH = 14 - 1.7 = 12.3


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