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| author | priyanka | 2015-06-24 15:03:17 +0530 |
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| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /3428/CH23/EX14.23.18/Ex14_23_18.sce | |
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initial commit / add all books
Diffstat (limited to '3428/CH23/EX14.23.18/Ex14_23_18.sce')
| -rw-r--r-- | 3428/CH23/EX14.23.18/Ex14_23_18.sce | 21 |
1 files changed, 21 insertions, 0 deletions
diff --git a/3428/CH23/EX14.23.18/Ex14_23_18.sce b/3428/CH23/EX14.23.18/Ex14_23_18.sce new file mode 100644 index 000000000..a19fbf110 --- /dev/null +++ b/3428/CH23/EX14.23.18/Ex14_23_18.sce @@ -0,0 +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+] +pH_1=-log10(C_1) +disp(pH_1,'pH of the required solution') +pH1=5 +C1=10^-5 //[H3O+] +pH2=3 +C2=10^-3 //[H3O+] +C_3=(C1+C2)/2 //[H3O+] +pH_2=-log10(C_3) +disp(pH_2,'pH of the required solution') + |