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R 2 = 0.94

5. R 2 = 0.94. q Cd = 9.58 C Cd 0.539. ln q Cd = 3.02 + 0.695 ln C Cd q Cd = 20.49 C Cd 0.695 q Cd = 9.58 C Cd 0.539. R 2 = 0.90. 12. σ H = q H – q OH = [SOH 2 + ] – [SO - ] – [(SO) 2 Ca 0 ] σ IS = 2[(SO) 2 Ca 0 ] – [SHPO 4 - ] σ OS = 0

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R 2 = 0.94

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  1. 5. R2 = 0.94 qCd = 9.58 CCd0.539

  2. ln qCd = 3.02 + 0.695 ln CCd qCd = 20.49 CCd0.695 qCd = 9.58 CCd0.539 R2 = 0.90

  3. 12. σH = qH – qOH = [SOH2+] – [SO-] – [(SO)2Ca0] σIS = 2[(SO)2Ca0] – [SHPO4-] σOS = 0 σD = -[SOH2+] + [SO-] + [SHPO4-]

  4. If SO- + H+ = SOH K1 SO- + 2H+ = SOH2+ K2 Develop log equations for K in terms of log cK and constant capacitance model surface phase activity coefficients. K1 = {[SOH] / [SO-] (H+)} exp (ΔzFΨ0 /RT) where Δz = 0 – (-1) = 1 logK1 = logcK + 2.303 (FΨ0 /RT) logcK + 2.303 (F2σP /SCRT) K2 = [SOH2+] / [SO-] (H+)2} exp (ΔzFΨ0 /RT) where Δz = 1 – (-1) = 2 logK2 = logcK + 2.303 (2FΨ0 /RT) logK2 = logcK + 2.303 (2F2σP /SCRT)

  5. Show that PZNPC = ½ logK2 K2 = [SOH2+] / [SO-] (H+)2} exp (ΔzFΨ0 /RT) At PZNPC, [SOH2+] = [SO-] and Ψ0 = 0 • K2 = 1 / (H+)2 So, log K2 = 2pH or PZNPC = ½ log K2

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