Question

Using matlab

The following system of five nonlinear equations gover the chemistry of rainwater ki 10$ (H) (100) Ky - ()(0,2 (HOO) Kw = (H')(OH) ©r b + (HCO, )+(co, 3). 0 = (HCO3)+2(CO3 ?) + (OH) + (H) where KH Henry's constant, K1, K2, and kw-equilibrium coefficients.cr total inorganic carbon (HCO3- bicarbonate (C032) - carbonate. (H') - hydrogen ion, and (OH)-hydroxyl ion Notice how the partial pressure of CO2 shows up in the equations indicating the impact of this greenhouse gas on the acidity of rain. Use these equations and the fsolve function to compute the pH of rainwater given that KH-10-146.44-10-63.42-10-10.3 and Kw 10714 Compare the results in 1958 when the P, was 315 and in 2015 when it was about 400 ppm. Note that this is a difficult problem to solve because the concentrations tend to be very small and vary over many orders of magnitude. Therefore, it is useful to use the trick based on expressing the unknowns in a negative log scale, K- -log10(). That is the five unknowns, cr total inorganic carbon bicarbonate, carbonate, OH, THCO3)(C032, and crean be reexpressed as the unknowns pH, pol pHCO3. PCO3, and perasin (H') - 10 (OH) - 10 H (HCO ) 10 (co, ) = 100 or 10 In addition, it is helpful to use optimset to set a stringent citeron for the function tolerance as in the following script you can use to generate your solutions cle, format compact xguess - 17;7;3;7; 3); options - optinset('tolfun, le-12) [x1, fxi) - fsolve(@funpt,xguess,options, 315); [x2, fx2] - fsolve(funpH, guess, options, 400); x1', fx1,x2',fx2 (Round the final answers to four decimal places)

Answer 1

There are following equilibrium for this:

1. CO₂ (aq) + H₂O    $\rightleftharpoons$ H₂CO₃ (aq)

2. H₂CO₃ (aq) $\rightleftharpoons$ H⁺ (aq) + HCO₃^- (aq) : K1 =10^-6.3

3. HCO₃^- (aq)  $\rightleftharpoons$ H⁺ (aq) + CO₃^2- (aq) : K2 =10^-10.3

4. H₂O + H₂O    $\rightleftharpoons$ H₃O⁺ + OH^- : Kw =10^-14

In 1958 ,   

pCO₂ = 315 ppm ,  

• we have K1 =10^-6.3 = 10^{6 *}[ H⁺ ][HCO₃^-  ] / K_H*pCO₂

on putting values,  10^-6.3 = 10^{6 *}[ H⁺ ][HCO₃^-  ] / 10^-1.46 *315

we have ,  [ H⁺ ] = [HCO₃^-  ] = 2.4*10^-6 M

• From  HCO₃^- (aq)  $\rightleftharpoons$ H⁺ (aq) + CO₃^2- (aq)   : K2 =10^-10.3

K2 = [CO₃^2-] [ H⁺] / [ HCO₃^-] =10^-10.3

we use [HCO₃-  ] = 2.4*10^-6 M;

we get [CO₃^2-] = [ H⁺] = 1.1*10^-8 M

Total [ H⁺] = 2.41*10^-6 M ; so pH = 5.62

now, C_T = 1.09*10^-5 M + 2.4*10^-6 M + 1.1*10^-8 M = 1.33*10^-5 M

(OH^-) = Kw /[ H⁺] = 4.15*10^-9

• In 2015 ,   

  pCO₂ = 400 ppm ,  

• we have K1 =10^-6.3 = 10^{6 *}[ H⁺ ][HCO₃^-  ] / K_H*pCO₂

• on putting values,  10^-6.3 = 10^{6 *}[ H⁺ ][HCO₃^-  ] / 10^-1.46 *400

  we have ,  [ H⁺ ] = [HCO₃^-  ] = 2.64*10^-6 M

• From  HCO₃^- (aq)  $\rightleftharpoons$ H⁺ (aq) + CO₃^2- (aq)   : K2 =10^-10.3

• K2 = [CO₃^2-] [ H⁺] / [ HCO₃^-] =10^-10.3

  we use [HCO₃-  ] = 2.64*10^-6 M;

  we get [CO₃^2-] = [ H⁺] = 1.15*10^-8 M

  Total [ H⁺] = 2.65*10^-6 M ; so pH = 5.58

  now, C_T = 1.39*10^-5 M + 2.64*10^-6 M+ 1.15*10^-8 M = 1.65*10^-5 M

  (OH^-) = Kw /[ H⁺] = 3.77*10^-9