10 mL of 0.10 M HCl is being titrated with 0.10 M NaOH. What would the pH of the solution be after addition of a) 1.0 mL of NaOH and b) 9.0 mL of NaOH
Both HCl and NaOH undergo ionization completely since they are strong acid and base respectively.
The reaction between NaOH and HCl can be represented as
NaOH+ HCl----> NaCl+ H2O (1 mole of HCl reacts with 1 mole of NaOH)
moles in 0.1M containing 10ml =0.001 moles
a) moles of NaOH in 0.1M and 1 ml= 0.0001 moles
HCl is excess and all the NaOH gets consumed
So moles of HCl remaining =0.001-0.0001=0.0009
Total volume after mixing = 10 (HCl)+ 1ml( NaOH)= 11ml
Molarity of HCl= 0.0009*1000/11=0.0818
HCl----> H+ Cl-
b) Moles of NaOH in 0.1M and 9ml =9*0.1/1000=0.0009
HCl is excess and all the NaLOH reacts and HCl remaining =0.001-0.0009=0.0001
Volume after mixing = 10+9=19
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