displaying values in command window

hi, trying to display some values but am not sure how to go about it, this is a pH array, Vb is volume of base added. What i am trying to do is display the midpoint to equivalence which is the half of Vb_ev and then display the corresponding pH value at midpoint, which i know is approx 4.756. see on from line 6 from the bottom.

      format short
      Kw = 1e-14;
      Ka = 1.755388e-5;
      pKa = 4.756;
      %
      Ca = 0.5;
      Cb = 0.1;
      Va = 100;
      Vb = 0.05:0.05:2000;
      Ma = (Ca * Va) / 1000;
      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);
            Vb_ev = Vb(i); 
            pH_ev = pH(end); 
        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
      % 
      Vb_Mid_Pt1 = (Vb_ev)*0.5;
      pH_Mid_Pt1 = i want to display pH at Vb_Mid_Pt1;
      disp('The Volume is, mL:')
      disp(Vb_Mid_Pt1)
      disp('The pH is:')
      disp(pH_Mid_Pt1)

[David Sanchez]

use the {}Code button to properly present your code, please.

[Jan]

"No luck" is a not a good description of the problem and of what you actually want. The less the readers have to guess, the more likely is a matching answer.

Answer 1

[kedija]

the reason you are getting such error is the index Vb_Mid_Pt1 is not integer. if u cast that it works fine. here you go

format short
      Kw = 1e-14;
      Ka = 1.755388e-5;
      pKa = 4.756;
      %
      Ca = 0.5;
      Cb = 0.1;
      Va = 100;
      Vb = 0.05:0.05:2000;
      Ma = (Ca * Va) / 1000;
      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);
            Vb_ev = Vb(i); 
            pH_ev = pH(end); 
        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
      % 
      Vb_Mid_Pt1 = uint16(Vb_ev)*0.5;
      pH_Mid_Pt1 = pH(Vb_Mid_Pt1);%i want to display pH at Vb_Mid_Pt1;
      disp('The Volume is, mL:')
      disp(Vb_Mid_Pt1)
      disp('The pH is:')
      disp(pH_Mid_Pt1)

[harley]

thanks what does the unit16 actual mean?

[David Sanchez]

It transforms the variable into Unsigned INTeger of 16 bits

[harley]

thanks kedija, im trying to get it to return the corresponding pH value when;

Vb = Vb_env*0.5

both these values are in column 5000. I get a pH value of 3.1646 which corresponds to column 250, which is the value obtained from;

Vb_Mid_Pt1 = uint16(Vb_ev)*0.5;

The pH should be around 4.756. I think my problem lies with the above expression.

thanks in advance.

[kedija]

I don't really know what your code does but Vb_ev is 500 and it is giving you the 250th value. if you want to display the 5000 value change ur code somehow to correct ur index. pH(5000) is 4.7556 .... I think you are messing something in the index.

Answer 2

[David Sanchez]

I hope this is what you want to do:

my_var = 3;

fprintf('The value of my variable is : %g \n' ,my_var);

OUTPUT:

The value of my variable is : 3