Homework Answers
NaCLO + HCl _----------> HClO + NaCl
55x 0.2 0 0 0 initial mmoles
1) before tittration
pH of a salt of weak acid and strong base is given by
pH = 1/2 [pKw + pKa + log C]
= 1/2[ 14 +(8-log3.5) + log 0.2]
= 10.3785
2) Half equivalence point
[NaClO] = [HclO] and the ph of the buffer is given by Hendersen equation
pH = pKa + log [conjugate base]/[acid]
= pKa
= 7.456
3) At the equivalence
0 0 11 11 after mmoles
0 0 11/110 concentrations
The solution has only weak acid , whose pH is calculated as
pH = 1./2 [pKa -logC]
=1/2[7.456 - log 0.1]
= 3.228
4) exces Hcl added
0 1.2 11 11 mmoles after reaction
the solution has a strong acid and a weak acid. Due to common ion effect all [H+] is from strong acid only .
Thus [H+] = 1.2/116 =0.0103 M
pH = - log 0.0103
= 1.985
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