1 files changed, 7 insertions, 6 deletions
diff --git a/3856/CH11/EX11.5/Ex11_5.sce b/3856/CH11/EX11.5/Ex11_5.sce
index 150d241e3..751cef6e4 100644
--- a/3856/CH11/EX11.5/Ex11_5.sce
+++ b/3856/CH11/EX11.5/Ex11_5.sce
@@ -1,3 +1,4 @@
+
 //Describe how you would prepare a phosphate buffer with a pH of seven point four
 
 //Example 11.5 
@@ -6,21 +7,21 @@ clc;
 
 clear;
 
-Ka1=7.5*10^-3; //Equilibrium consatnt for H3PO4= H+ +H2PO4-
+Ka1=7.5*10^-3; //Equilibrium constant for H3PO4= H+ +H2PO4-
 
 pKa1=-log10(Ka1); //minus logerithm of Ka1
 
-Ka2=6.2*10^-8; //Equilibrium consatnt for H2PO4-= H+ +HPO4--
+Ka2=6.2*10^-8; //Equilibrium constant for H2PO4-= H+ +HPO4--
 
-pKa2=-log10(Ka2); //minus logerithm of Ka2
+pKa2=-log10(Ka2); //minus logarithm of Ka2
 
-Ka3=4.8*10^-13; //Equilibrium consatnt for HPO4-- = H+ +PO3---
+Ka3=4.8*10^-13; //Equilibrium constant for HPO4-- = H+ +PO3---
 
-pKa3=-log10(Ka3); //minus logerithm of Ka3
+pKa3=-log10(Ka3); //minus logarithm 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
+C1=10^(pH-pKa2); //Concentration 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