displaying values in command window

follow script, how do i display the value of Vb and pH when "abs(M_excess)<eps" and following that how would i plot that point a figure.

Kw = 1e-14;
Ka = 1.755388e-5;
pKa = 4.756;
%
Ca = 0.5;
Cb = 0.1;
Va = 100;
Vb = 0.05:0.05:2000;
%
% This simulates initial molarity of CH3CO2H.
Ma = (Ca * Va) / 1000;
% This simulates initial molarity of NaOH.
Mb = (Cb .* Vb) ./ 1000;
for i = 1:length(Mb) 
M_excess = Ma - Mb(i);
if abs(M_excess)<eps
Hplus = Ka * ((Ma_final * 0.999999) ./ Mb_final);
elseif M_excess > 0
  Ma_final = (M_excess * 1000) ./ (Va + Vb(i)); 
  Mb_final = (Mb(i) * 1000) ./ (Va + Vb(i));
  Hplus = Ka * (Ma_final ./ Mb_final);
elseif M_excess < 0
  OH = (M_excess * 1000 * (-1)) ./ (Va + Vb(i));
  Hplus = Kw ./ OH;
end
pH(i) = -log10(Hplus);
end

Answer 1

[Azzi Abdelmalek]

for i = 1:length(Mb)
  M_excess = Ma - Mb(i);
  v=[];
  if abs(M_excess)<eps
      Hplus = Ka * ((Ma_final * 0.999999) ./ Mb_final);
      v(end+1)=Hplus,
  elseif M_excess > 0
      Ma_final = (M_excess * 1000) ./ (Va + Vb(i));
      Mb_final = (Mb(i) * 1000) ./ (Va + Vb(i));
      Hplus = Ka * (Ma_final ./ Mb_final);
  elseif M_excess < 0
      OH = (M_excess * 1000 * (-1)) ./ (Va + Vb(i));
      Hplus = Kw ./ OH;
  end
  pH(i) = -log10(Hplus);
end
display(v)

[harley]

hello again, in my command window all it displays is

v =

     []

[Azzi Abdelmalek]

Which means the condition is not satisfied

[harley]

thanks for your time again again Azzi, i am a complete novice with matlab, any other thoughts on displaying Vb and pH at this point?

[Azzi Abdelmalek]

display(Vb')
display(pH')

% or you can display them together

[Vb' pH']

[harley]

i think i got it to work Azzi. Thanks for your help, its been great.

for i = 1:length(Mb)
M_excess = Ma - Mb(i);
if abs(M_excess)<eps
    Hplus = Ka * ((Ma_final * 0.999999) ./ Mb_final);
    pH_ev = pH(end);
    Vb_ev = Vb(i);
elseif M_excess > 0
    Ma_final = (M_excess * 1000) ./ (Va + Vb(i));
    Mb_final = (Mb(i) * 1000) ./ (Va + Vb(i));
    Hplus = Ka * (Ma_final ./ Mb_final);
elseif M_excess < 0
    OH = (M_excess * 1000 * (-1)) ./ (Va + Vb(i));
    Hplus = Kw ./ OH;
end
pH(i) = -log10(Hplus);
end
% OUTPUT SECTION
disp('The Volume of Base NaOH at Equivalence is:')
disp(Vb_ev)
disp('The pH at Equivalence is:')
disp(pH_ev)