www.sapinglearning.comibiscms/modiresource/view.php?inpopup-tr R: Topic 7 Skills

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www.sapinglearning.comibiscms/modiresource/view.php?inpopup-tr R: Topic 7 Skills Practic... Jump to... ring 17 MCGREGOR I Activities and Due Dates I Topic 7 Skills Practice Video E Grade 3/5/2017 11:55 PM o13 Print Calculator Periodic Table stion 2 of 3 Map Sapling Learning The pKa of hypochlorous acid is 7530. A 58.0 mL solution of0.140 M sodium hypochlorite (Naoc) is titrated with 0.290 M HCI. Calculate the pH of the solution a) after the addition of 10.1 mL of 0.290 M HCI. pH b) after the addition of 29.8 mL of 0.290 M HCI. pH c) at the equivalence point with 0.290 M Hca. Number pH A Previous Give up & view soluion 2 check Answer 0Ned Exit about us careers partners privacy policy terms of use contac

Explanation / Answer

a)

millimoles of NaOCl = 58 x 0.140 = 8.12

millimoles of HCl = 10.1 x 0.290 = 2.929

OCl- + HCl --------------> HOCl + Cl-

8.12 2.929 0 0

5.191 0 2.929   

pH = pKa + log [salt / acid]

= 7.530 + log [5.191 / 2.929]

= 7.78

pH = 7.78

b)

millimoles of HCl = 29.8 x 0.290 = 8.642

OCl- + HCl --------------> HOCl + Cl-

8.12 8.642 0 0

0 0.522 8.12

here stong acid reamins.

so [H+] = 0.522 / (58 + 29.8) = 5.95 x 10^-3 M

pH = -log [H+] = -log (5.95 x 10^-3)

pH = 2.23

c)

at equivalence point

volume = 8.12 / 0.290 = 28

salt concentration = 8.12 / (58 + 28) = 0.0944

pH = 7 - 1/2 (pKa + log C)

= 7 - 1/2 (7.530 + log 0.0944)

= 3.75

pH = 3.75

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