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//Describe how you would prepare a phosphate buffer with a pH of seven point four
//Example 11.5
clc;
clear;
Ka1=7.5*10^-3; //Equilibrium consatnt for H3PO4= H+ +H2PO4-
pKa1=-log10(Ka1); //minus logerithm of Ka1
Ka2=6.2*10^-8; //Equilibrium consatnt for H2PO4-= H+ +HPO4--
pKa2=-log10(Ka2); //minus logerithm of Ka2
Ka3=4.8*10^-13; //Equilibrium consatnt for HPO4-- = H+ +PO3---
pKa3=-log10(Ka3); //minus logerithm of Ka3
pH=7.40; //pH of the required buffer solution
C1=10^(pH-pKa2); //Concentratin of required solution to prepare buffer solution of pH of 7.40
C=C1/1.0; //Ratio of the required solution to prepare buffer solution of pH of 7.40
printf("Ratio of the required solution = %.2f The buffer is dissolve to disodium hydrogen phosphate and sodium dihydrogen phosphate in a mole ratio of 1.5:1.0 ",C);
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