Question

When 10.00 mL of 0.240 M HX (Ka = 4.31 10-5) is titrated with 0.120 M...

When 10.00 mL of 0.240 M HX (Ka = 4.31 multiply.gif 10-5) is titrated with 0.120 M KOH, the pH will increase. Calculate the pH of the solution at each point in the titration. The pKw is 14.000 at this temperature.

Volume of KOH (mL) pH
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
16.00
18.00
20.00
22.00
24.00
26.00
28.00
0 0
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Answer #1

6.24 M . [salt] e solution 10ml w HX + KOH KX + H₂O 2.4 milindes 0.12M So where Untill 20 ml of kon is added there will be wevolume tme) • 10 pn = 4.36 + log [1:2] • 12 → pu- 4.36 + lug [1.44] = 4.536 To:96] 1.14 P 1.16 . 18 - ph = 4.36 + log [ 1.68]volume 24 aml of Nokon o left [ou] = ruxo 127 = opona -log(48012] ч РИs - go 12.15 - 1.85 86 %) Гбх6-12-12 fоr -- т с ко - (

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