saplinglearning.com ons - CHEM 202-Fall17- McMAHON: HW 14-II Strong Acids/Bases
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saplinglearning.com ons - CHEM 202-Fall17- McMAHON: HW 14-II Strong Acids/Bases Jump to Map University Science Books presented by Sapling Leaming General Chemistry 4th Edition Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.160 M HCIO(aq) with 0.160 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 mL of KOH Number (c) after addition of 35.0 mL of KOH Number (d) after addition of 50.0 mL of KOH Number (e) after addition of 60.0 mL of KOH O Previous Give Up & View Solution check Answer 0 Next -I Exit Hint about us careers privacy policy terms of use contact us help 29 MacBook 80Explanation / Answer
millimoles of HClO= 50 x 0.160= 8
pKa = -logKa = = -log(4.0 x10^ -8.) = 7.4
a) before additon of KOH added
pH = 1/2 (pKa- log )
= 1/2 (7.4 -log (0.160) ) = 4.10
pH= 4.10
(b) after addition of 25.0 mL of KOH
it is first equivalece point here pH = pKa
pH = 7.40
(c) after addition of 35.0 mL of KOH
millimoles of KOH = 35 x 0.160 = 5.6
HClO + KOH ------------------------------> KClO + H2O
8 5.6 0 0 -----------------------initial
2.4 0 5.6 5.6----------------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 (5.6 / 2.4)
= 7.77
pH = 7.77
(d) after addition of 50.0 mL of KOH
millimoles of KOH = 0.160 x 50 = 8
HClO + KOH ------------------------------> KClO + H2O
8 8 0 0 -----------------------initial
0 0 8 8------------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
= 8 /(50+50)
= 0.08 M
pH = 7 + 1/2[Pka + logC]
= 7 + 1/2 [7.4 + log (0.08)]
= 10.15
pH = 10.15
e) after addition of 60.0 mL of KOH
pH = 12.16
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