Answer to Question #224840 in Biochemistry for MAGCABA

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) 

Expert's answer

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"



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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