VolHX Vol OH. Total rX) Init, mL added for th Init., mL increment, mLI mL s ! Vo

Question

VolHX Vol OH. Total rX) Init, mL added for th Init., mL increment, mLI mL s ! VolOH, lTotalVol [HKLa | [X]/HX) | log Soin, mL 50.0 0.0 0.0 50.0 | 100 | 600 14% b.D33 | 0.050| 1.50 | . I7lr soo 10.0 21 50.0 | lo.ott loo31 10.04e| 1:45 | 1140 |4.Yo 15.0 | 65.0 5.0 50.0 50.0 | 10.0 3bD | IOD-D ,65b | 6,020 0.03p | l. 50 | | 5.2 g th | 4D-0 | |y6.D | 0.D30 | 0, D lH | 0. Daa| 159 | . Ique | S-45 5-5 450 I95,o bo210-01 LO.6Up 11.45 | .lul 50.0 5,US | 4. 45s 5 5445 5-4 50-55 0-0. Ou 001 3-3-2-0|D D-O-O' 00 3-2 0-0 0006 Vol D-5 0-1 2-3 4 5 f the tenitation Constant of a Weak Acid - 45

Explanation / Answer

Calculation of pKa

Sample : 2

moles = molarity x volume

So,

moles of acid present (HX) = 0.1 M x 50 ml = 5 mmol

moles of NaOH added = 0.1 M x 10 ml = 1 mmol

[NaX] formed = 1 mmol/60 ml = 0.0167 M

[HX] remained = 4 mmol/60 ml = 0.0667 M

this is a buffer solution

Using,

pH = pKa + log(base/acid)

with,

[base] = NaX

[acid] = HX

pH = 4.76

we get,

pKa = pH + log(NaX/HX)

       = 4.76 + log(0.0167/0.0667)

       = 4.16

Similarly calculations from other sample data can be done.

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.