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)
when 1 mol 0.1M HCl and 3 mol of 0.5 M NaOH are added
milli-mole of HCl=1
millimole of
0.1 mole of HCl is neutralized by 1.5 mole NaOH
Remaining mole of NaOH =1.5-0.1=1.4
0.4 0.4 0.4
remaining millimole of NaOH =1.4-0.4=1.0
concentration of NaOH, [NaOH}=
PH of resulting solution after addition of 1mol of 0.1 M HCl and 3ml of 0.5M NaOH is 12.89
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