Question

How is the pH changed on adding 1 ml of 0.1 mol/litre HCl followed by 3ml of 0.5 mol/litre NaOH to the mixture in question 6 above? (Show your answer by means of calculation taking consideration of transitions taking place and showing how the pH of the solutions is affected) 

Answer 1

when 1 mol 0.1M HCl and 3 mol of 0.5 M NaOH are added

$HCl+NaOH\to NaCl+H_2O$

milli-mole of HCl=1$\times 0.1=0.1$

millimole of $NaOH =0.5\times 3=1.5$

0.1 mole of HCl is neutralized by 1.5 mole NaOH

Remaining mole of NaOH =1.5-0.1=1.4

$CH_3COOH+NaOH \to CH_3COONa+H_2O$

0.4 0.4 0.4

remaining millimole of NaOH =1.4-0.4=1.0

concentration of NaOH, [NaOH}=$1/(9+1+3)=1/13=0.077$

$POH=-log(0.077)=1.11$

$PH=14-1.11=12.89$

PH of resulting solution after addition of 1mol of 0.1 M HCl and 3ml of 0.5M NaOH is 12.89