Please help and be as descriptive as possible on how you solve, especially with
ID: 508744 • Letter: P
Question
Please help and be as descriptive as possible on how you solve, especially with the ICE Table so it's easy for me to follow.
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.130 M HCIO(ag) with 0.130 M KOHaq). 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
millimoles of HClO= 50 x 0.130= 6.5
pKa = -logKa = = -log(4.0 x10^ -8.) = 7.4
a) before additon of KOH added
pH = 1/2 (pKa- log C)
= 1/2 (7.4 -log (0.130) ) = 4.14
pH= 4.14
(b) after addition of 25.0 mL of KOH
it is first equivalece point here pH = pKa
pH = 7.40
(c) after addition of 30.0 mL of KOH
millimoles of KOH = 30 x 0.130 = 3.9
HClO + KOH ------------------------------> KClO + H2O
6.5 3.9 0 0 -----------------------initial
2.6 0 3.9 ------------------equilibirum
pH = pKa + log[salt/acid]
= 7.4 + log (3.9 / 2.6)
= 7.58
pH = 7.58
(d) after addition of 50.0 mL of KOH
This is equivalence point. here salt only remains.
it is the salt of strong base and weak acid so pH should be more than 7
[salt] = salt millimoles /total volume in ml
= 6.5 /(50+50)
= 0.065 M
pH = 7 + 1/2[Pka + logC]
= 7 + 1/2 [7.4 + log (0.065)]
= 10.11
pH = 10.11
e) after addition of 60.0 mL of KOH
millimoles of KOH = 0.130 x 60 = 7.8
in the solution strong base remained
[base ] = 1.3 /total volume = 1.3 /110 = 0.0118
pOH = -log[OH-] = -log( 0.0118) =1.93
pH + pOH = 14
pH = 12.07