calculate the pH for each of the following cases in the titration of 50.0 mL of
ID: 1027248 • Letter: C
Question
calculate the pH for each of the following cases in the titration of 50.0 mL of 0.170 M HClO (aq) with 0.170 M KOH (aq). The ionization constant for HClO can be found here (4.0*10^-8)
General Chemistry 4th Edition University Science Books presented by Sapling Leaming Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.170 M HCIo(aq) with 0.170 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 30.0 mL of KOH Number (d) after addition of 50.0 mL of KOH Number (e) after addition of 60.0 mL of KOHExplanation / Answer
pKa = -logKa = = -log(4.0 x10^ -8.) = 7.4
millimoles of HClO= 50 x 0.170= 8.5
a) 0 ml KOH added
pH = 1/2 (pKa- log C)
= 1/2 (7.4 -log (0.170) ) = 4.08
pH= 4.08
(b) after addition of 25.0 mL of KOH
it is first equivalece point here pH = pKa
pH = 7.4
(c) after addition of 30.0 mL of KOH
millimoles of KOH = 30 x 0.170 = 5.1
HClO + KOH ------------------------------> KClO + H2O
8.5 5.1 0 0 -----------------------initial
3.4 0 5.1 5.1 -------------------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 (5.1 /3.4)
= 7.58
pH = 7.58
(d) after addition of 50.0 mL of KOH
millimoles of KOH = 0.170 x 50 = 8.7
HClO + KOH ------------------------------> KClO + H2O
8.5 8.5 0 0 -----------------------initial
0 0 8.5 8.5----------------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
= 8.5 /(50+50)
= 0.085 M
pH = 7 + 1/2[Pka + logC]
= 7 + 1/2 [7.4 + log (0.085)]
= 10.16
pH = 10.16
e) after addition of 60.0 mL of KOH
pH = 12.19
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