Question #224840

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) 


1
Expert's answer
2021-10-19T03:45:02-0400

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

HCl+NaOHNaCl+H2OHCl+NaOH\to NaCl+H_2O

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

millimole of NaOH=0.5×3=1.5NaOH =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

CH3COOH+NaOHCH3COONa+H2OCH_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=0.0771/(9+1+3)=1/13=0.077

POH=log(0.077)=1.11POH=-log(0.077)=1.11

PH=141.11=12.89PH=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



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