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//Section-14,Example-1,Page no.-PC.112
//To calculate the pH values of the following.
clc;
C_HCl=0.001
C_1=C_HCl //Since HCl is a strong acid,[H3O+]=[HCl]
pH_1=-log10(C_1)
disp(pH_1,'pH of 0.001 M HCl')
C_NaOH=0.0001
C_2=C_NaOH //Since NaOH is a strong base so [OH-]=[NaOH]
pOH=-log10(C_2)
pH_2=14-pOH
disp(pH_2,'pH of 0.0001 M NaOH')
C_BaOH2=0.001
C_31=2*C_BaOH2 // [OH-]=2*[Ba(OH)2]
k_w=10^-14
C_32=k_w/(C_31)
pH_3=-log10(C_32)
disp(pH_3,'pH of 0.001 M Ba(OH)2')
M=(0.049/98)/(200/1000) //Molarity of H_2SO_4 solution
C_4=2*M //[H_3O+]
pH_4=-log10(C_4)
disp(pH_4,'pH of the given solution')
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