If someone could please help me finish up this problem I would REALLY appreciate
ID: 870950 • Letter: I
Question
If someone could please help me finish up this problem I would REALLY appreciate it! Thank you!!
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.200 M HCIO(aq) with 0.200 M KOH(aq), The ionization constant for HCIO can be found here. Number (a) before addition of any KOH 405 (b) after addition of 25.0 mL of KOH 7.146 (c) after addition of 30.0 mL of KOH 7.276 (d) after addition of 50.0 mL of KOH 102 (e) after addition of 60.0 mL of KOH With equal concentrations of monoprotic titrant and analyte, the equivalence point would occur when the added volumes are equal (50.0 mL of KOH added). So at point (b), we are 25.0/50.0 = 50% of the way to the equivalence point. If 50% of the acid has reacted, then 50% remains, and 50% of the conjugate base has been formed so [CIO^-][HCIO] 50/50 = 1. Find the pH using the Henderson-Hasselbalch equation.Explanation / Answer
HOCl , Ka value = 4.0 x10^ -8
pKa = -logKa = = -log(4.0 x10^ -8.) = 7.4
millimoles of HClO= 50 x0.2= 10
a) before addition any KOH added
pH = 1/2 (pKa- log C)
= 1/2 (7.4 - log (0.2) ) = 4.05
pH= 4.05
(b) after addition of 25.0 mL of KOH
millimoles of KOH = 0.2 x 25 = 5
HClO + KOH ------------------------------> KClO + H2O
10 5 0 0 -----------------------initial
5 0 5 5-------------------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 (5/5)
pH = 7.4
(c) after addition of 30.0 mL of KOH
millimoles of KOH = 0.2 x 30 = 6
HClO + KOH ------------------------------> KClO + H2O
10 6 0 0 -----------------------initial
4 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 (6/4)
= 7.57
pH = 7.57
(d) after addition of 50.0 mL of KOH
millimoles of KOH = 0.2 x 50 = 10
HClO + KOH ------------------------------> KClO + H2O
10 10 0 0 -----------------------initial
0 0 10 10 -------------------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
= 10/(50+50)
= 0.1 M
pH = 7 + 1/2[Pka + logC]
= 7 + 1/2 [7.4 + log (0.1)]
= 10.2
pH = 10.2
e) after addition of 60.0 mL of KOH
millimoles of KOH = 0.2x 60 = 12
HClO + KOH ------------------------------> KClO + H2O
10 12 0 0 -----------------------initial
0 2 10 10-------------------equilibirum
in the solution strong base remained
[base ] = 2/total volume = 2/110 = 0.018
pOH = -log[OH-] = -log(0.018) =1.74
pH + pOH = 14
pH = 12.26
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