Index them!
Since all the ‘i’ vectors are the same, a few minor corrections (and some repetitive editing) produces:
iv=0:0.1:1;
%chain 1
for k = 1:numel(iv)
i = iv(k);
f_a=i;
f_b=-i+1;
r_a1=1;
r_b1=1;
F_a1(k)=(((r_a1.*((f_a).^2))+(f_a.*f_b))/((r_a1.*((f_a).^2))+(2.*f_a.*f_b)+(r_b1.*((f_b).^2))));
end
%chain 2
for k = 1:numel(iv)
i = iv(k);
f_a=i;
f_b=-i+1;
r_a2=0.25;
r_b2=0.25;
F_a2(k)=(((r_a2*((f_a).^2))+(f_a*f_b))/((r_a2*((f_a).^2))+(2*f_a*f_b)+(r_b2*((f_b).^2))));
end
%chain 3
for k = 1:numel(iv)
i = iv(k);
f_a=i;
f_b=-i+1;
r_a3=4;
r_b3=4;
F_a3(k)=(((r_a3*((f_a).^2))+(f_a*f_b))/((r_a3*((f_a).^2))+(2*f_a*f_b)+(r_b3*((f_b).^2))));
end
%chain 4
for k = 1:numel(iv)
i = iv(k);
f_a=i;
f_b=-i+1;
r_a4=0.25;
r_b4=4;
F_a4(k)=(((r_a4*((f_a).^2))+(f_a*f_b))/((r_a4*((f_a).^2))+(2*f_a*f_b)+(r_b4*((f_b).^2))));
end
figure
hold on
scatter(iv,F_a1,"black","filled")
scatter(iv,F_a2,"red","filled")
scatter(iv,F_a3,"blue","filled")
scatter(iv,F_a4,"green","filled")
title('F_A End Chain Versus f_a Monomer Available')
ylabel('F_A')
xlabel('Amount of f_a Monomer')
legend("Chain 1","Chain 2","Chain 3","Chain 4", 'Location','best')
hold off
That should do what you want.
