4/9/2018 11.55 PM 3 49/10 4/8/2018 1235 AM PintCalulatorPerlodic Table Question

Question

4/9/2018 11.55 PM 3 49/10 4/8/2018 1235 AM PintCalulatorPerlodic Table Question 16 of 20 Map General Chemistry 4th Edition University Science Books presented by Sapling Leaming Calculate the pH for each of the following cases in the titration of500 mL of 0210 M HCO ad with 0.210 M KOH(aq). The ionization constant for HCIO can be found here. (a) before addition of any KOH (b) after addition of 25.0 mL of KOH (e) after addition of 40.0m of KOH ld ather adsition of 50.0 ml of KOH O (e) after addition of 60.0 mL of KOH O Pre sous @Grve Up & View Sonton Q,Check An.wer?Next flExit ype here

Explanation / Answer

millimoles of HClO= 50 x0.210= 10.5

pKa = -logKa = = -log(4.0 x10^ -8.) = 7.4

a) before additon of KOH added

pH = 1/2 (pKa- log )

   = 1/2 (7.4 -log (0.210) ) = 4.04

pH= 4.04

(b) after addition of 25.0 mL of KOH

it is first equivalece point here pH = pKa

pH = 7.40

(c) after addition of 40.0 mL of KOH

millimoles of KOH = 40 x 0.210 = 8.4

HClO + KOH ------------------------------> KClO + H2O

10.5          8.4                                      0               0 -----------------------initial

2.1              0                                          8.4            8.4 ------------------equilibirum

pH = pKa + log[salt/acid]

    = 7.4 + log (8.4 / 2.1)

    = 8.00

pH = 8.00

(d) after addition of 50.0 mL of KOH

[salt] = salt millimoles /total volume in ml

           = 10.5 /(50+50)

           = 0.105 M

pH = 7 + 1/2[Pka + logC]

   = 7 + 1/2 [7.4 + log (0.105)]

    = 10.21

pH = 10.21

e) after addition of 60.0 mL of KOH

pOH = -log[OH-] = -log(0.0191) =1.72

pH + pOH = 14

pH = 12.28

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

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