diff options
Diffstat (limited to '3856/CH11/EX11.5')
| -rw-r--r-- | 3856/CH11/EX11.5/Ex11_5.sce | 13 |
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
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