0933 0914 6 0931 0912 S 0965 0929 907 035 Na.CCT\", CIO, 10, HCo, n,POH5O,.HAOs

Question

0933 0914 6 0931 0912 S 0965 0929 907 035 Na.CCT", CIO, 10, HCo, n,POH5O,.HAOs CoNH MNO .CH.Co.CIch C0j.(CHN 450 0.928 082 m 4 0926 0900 816 964 0924 .80 75 0.872 0.755 069 052 045 0064 OH F.SCN-,OCN-HS.CIO, Cios, BrO,.0.M0 350 300 964 0925 0.89305 0.355 CHACH CH,CO)(CH,CH.CH,C0h 700 0872 0755 0.685 050 0425 600 0.870 0.749075 0A5 0.405 Hcitrate- 450 0867 042 06 0A55 0.37 400 0857 0.740 0.660 0.445 0.355 Charge = AP", Fe, CSe, YIn citrate- lanthanides 900 500 0.728 0.51 0.405 018 0.115 400 0.725 0.50S 0.395 06 0095 0.738 0.34 0445 0.245 0.18 Charge. ±4 0588 035 0255 0.10 0.065 500 057 0.31 0.20 0048 002 1 100 Fe(CN V. Using activities, calculate the pH of 0.100 M HCIO2 (pka =1.96)

Explanation / Answer

Find the ionic strength first,
= 1/2 (CiZi2)
C is the concentration = 0.100 M and Z is the charge, for H+ it is +1 and for ClO2- it is -1
= 1/2 [ 0.1*(+1)2+0.1*(-1)2] = 0.1
At   = 0.1  H+ = 0.83 and ClO2- = 0.79
PH = - log ( [H+] H+) = -log [0.1*0.83] = 1.0809
Therefore PH = 1.08

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