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Fresno Pacific University x C Chegg.com C fi D WWW saplinglea rning com /ibiscms/mod/ibis/view.php?id-1496213 EE ppy Nernst Equation E... Engaging Students, Course: 15/USP-Blo... Google Docs ca Fresno Pacific Unive... go Turnitin StudentLoans.gov College GPA Calcula A Garmin Express Question 17 of 28 ncorrect 85 Map University Science Books General Chemistry presented by Sapling Learnin Donald McQuarrie Peter A. Rock. Ethan Gallogly 100 Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.150 M HCIO(ag) with 0.150 M KOH(aq). The ionization constant for HCIO can be found here. 38 Number 95 (a) before addition of any KoH 100 Number 100 (b) after addition of 25.0 mL of KOH 10 98 Tools 11 100 (c) after addition of 40.0 mL of KOH 12 95 13 98 Number (d) after addition of 50.0 mL of KOH 14 95 15 95 Number (e) after addition of 60.0 mL of KOH 16 98 17 A Previous 3 Give Up & View Solution Check Answer Next Exit Yc Yc uf Yc yc Yc 55 PM 5/4/2015Explanation / Answer
pKa = -logKa = = -log(4.0 x10^ -8.) = 7.4
millimoles of HClO = 50 x 0.15 = 7.5
a) before addition any KOH added
pH = 1/2 (pKa - log C)
= 1/2 (7.4 - log (0.15) ) = 4.1
pH= 4.1
(b) after addition of 25.0 mL of KOH
millimoles of KOH = 0.15 x 25 = 3.75
HClO + KOH ------------------------------> KClO + H2O
7.5 3.75 0 0 -----------------------initial
3.75 0 3.7 5 3.75 -------------------equilibirum
in the solution acid and salt remained so it can form buffer
For acidic buffer
Henderson-Hasselbalch equation
pH = pKa + log[salt/acid]
= 7.4 + log (3.75/3.75)
pH = 7.4
(c) after addition of 40.0 mL of KOH
millimoles of KOH = 0.15 x 40 = 6
HClO + KOH ------------------------------> KClO + H2O
7.5 6 0 0 -----------------------initial
1.5 0 6 6 -------------------equilibirum
in the solution acid and salt remained so it can form buffer
For acidic buffer
Henderson-Hasselbalch equation
pH = pKa + log[salt/acid]
= 7.4 + log (6/1.5)
= 8.00
pH = 8.00
(d) after addition of 50.0 mL of KOH
millimoles of KOH = 0.15 x 50 = 7.5
HClO + KOH ------------------------------> KClO + H2O
7.5 7.5 0 0 -----------------------initial
0 0 7.5 7.5-------------------equilibirum
in the solution salt remained so we have to use salt hydrolysis.
it is the salt of strong base and weak acid so pH should be more than 7
[salt] = salt millimoles /total volume in ml
= 7.5/(50+50)
= 0.075 M
pH = 7 + 1/2[Pka + logC]
= 7 + 1/2 [7.4 + log (0.075)]
= 10.13
pH = 10.13
e) after addition of 60.0 mL of KOH
millimoles of KOH = 0.15 x 60 = 9
HClO + KOH ------------------------------> KClO + H2O
7.5 9 0 0 -----------------------initial
0 1.5 7.5 7.5-------------------equilibirum
in the solution strong base remained
[base ] = 1.5/total volume = 1.5/110 = 0.0136
pOH = -log[OH-] = -log(0.0136) =1.86
pH + pOH = 14
pH = 12.14
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