updated the code

1 files changed, 11 insertions, 11 deletions

diff --git a/3428/CH23/EX14.23.18/Ex14_23_18.sce b/3428/CH23/EX14.23.18/Ex14_23_18.sce
index a19fbf110..0fdda2abd 100644
--- a/3428/CH23/EX14.23.18/Ex14_23_18.sce
+++ b/3428/CH23/EX14.23.18/Ex14_23_18.sce
@@ -1,21 +1,21 @@
 //Section-14,Example-2,Page no.-PC.112
 //To calculate the pH in the following cases.
 clc;
-V_1=150                //volume of 0.1 NaOH solution
-V_2=150                //volume of 0.2 HCl solution
-N_1=0.1
-N_2=0.2
-V=V_1+V_2             //Total volume of the solution
-m_eq=(V_2*N_2)-(V_1*N_1)            //Total milliequivalents of excess HCl
-N=m_eq/V
-C_1=N                          //Since HCl is a strong acid so[HCl]=[H3O+]
+V_1=150                //volume of 0.1 NaOH solution(ml)
+V_2=150                //volume of 0.2 HCl solution(ml)
+N_1=0.1                 //(N)
+N_2=0.2                    //(N)
+V=V_1+V_2             //Total volume of the solution(ml)
+m_eq=(V_2*N_2)-(V_1*N_1)            //Total milligram equivalents of excess HCl(gm equivalents)
+N=m_eq/V                              //(N)
+C_1=N                          //Since HCl is a strong acid so[HCl]=[H3O+]   (M)
 pH_1=-log10(C_1)
 disp(pH_1,'pH of the required solution')
 pH1=5
-C1=10^-5                    //[H3O+]
+C1=10^-5                    //[H3O+]             (M)
 pH2=3
-C2=10^-3                    //[H3O+]
-C_3=(C1+C2)/2                //[H3O+]
+C2=10^-3                    //[H3O+]                (M)
+C_3=(C1+C2)/2                //[H3O+]                  (M)
 pH_2=-log10(C_3)
 disp(pH_2,'pH of the required solution')