Question

_{The pk, of hypochlorous acid is 7.530. A 59.0 mL solution of 0.143 M sodium hypochlorite (NaOCl) is titrated with 0.264 MHCI. Calculate the pH of the solution after the addition of 9.91 mL of 0.264 M HCI. pH = Calculate the pH of the solution after the addition of 34.2 mL of 0.264 M HCI. pH = Calculate the pH of the solution at the equivalence point with 0.264 M HCI. pH =}

Answer 1

ans, Given The pha hypo chlorous acid is 7.530 59.0 m2 solution of 0143 M Sodium hypochlorite . (Naoel) Hl= 0.264M Mill moles of Naod. = 59 x 0.143 = 8.437 e After the addition of 9.91 mb of 0.264 = 9.91 X 0264 M = 2.6162 mill moles of Naod = 8.437– 2.6162 = 5.8208 millimoles of Naoch left 2.6162 milli moles Hod will from

[nlaocl) = 5.8208 67.89 0857 (stad) = 0.0854) [Hoc ) = 2.6162 67.89 ( [Hocl) = 0.0385 explot = pkan t dog ocean ospite = 7.5304 og 12.2259) << pH = 7.530+ 0.3475 pH = 7.87 pH = 7.87 After the addition of 34.2 m2 of 0.264 mud millimales of Hocl = (34.2) (0.264)

= 9.0288 --> 9.0288 – 8.437 = 0.5918 millimoles of hd left Chcd ) = 0.5918 85.3 = 6.93x10-30 [HI] = 6.9381" 3M As wel ie strong acid (ncl)• Cory ple = -blog forty pH = -log (693x103) =-(-2.15) pe = 215 © At Equivalent point all Naoch becomes Holl 8.434 = V(0.32) V= 8.437 0.32 V=26,36 m2 Total volume 559 +26:36 = 85.36 mL

[hc] = 8.437 85.36 [HI] = 0.098m pH = + 2 [pka_ loge =* (4.530 -dog (0.0el -4 (7530(-1.0083] =4 (7.530+1-0087] = 18.5387) = 4.26 i [p= 4126) سے۔ Date meiner
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