Model: I(total ions) = I(H+) + I(HCl+) = A(t) P m I(H+) = A(H+) P n I(H+)/(I(H+)+I(HCl+)) = (A(H+)/A(t)) P (n-m) = A(H+,t) P (n-m) ln(I(H+)/(I(H+)+I(HCl+))) = (n-m) ln(P) + ln(A) HCl, F<-<-X, (1,0), J = 6, mass analysis: agust,

pnt we04awe06awe09awe13a ,653331, ,752685, ,952699,629396H ,753166, ,953329, , ,253727,529609, ,450176, ,5HCl ,717501,6-4797, ,134539, ,146695,419527, ,250727, ,653331,630513bgr.(H+) ,253727,530691bgr.(HCl+) ,2-1444,7-2717,5I(H+) , , I(HCl+) 0, , , ,023994I(H+)/(I(H+)+I(HCl+)) -4, , , ,72995ln(ratio) 0,0350,060,090,13power (after autotracker) -3, , , ,04022ln(P)

Coefficient values ± one standard deviation a (inters ) = ± b(slope) = ± ln(I(H+)/(I(H+)+I(HCl+))) ln(P)

Coefficient values ± one standard deviation a = ± b = ± ln(I(H+)/(I(H+)+I(HCl+))) ln(P)

Coefficient values ± one standard deviation a = ± Inf b = ± Inf ln(I(H+)/(I(H+)+I(HCl+))) ln(P)

Summary: ln(P) ln(I(H+)/(I(H+)+I(HCl+))) (n-m) = 0.79 (n-m) = 0.69 (n-m) = 0.63

Conclusion: Slope is resonably close to the value 1, i.e. (n-m) = 1 as to be expected for the case n = 4 and m = 3