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What change in pH should be observed if 10.0 mL of 0.100 M NaOH is added...

What change in pH should be observed if 10.0 mL of 0.100 M NaOH is added to 100 mL of a buffer that is 0.100 M in CH3COOH and 0.100 M in NaCH3CO2? Ka for acetic acid is 1.8 x 10-5.

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Answer #1

Step -1

Apply Handerson's equation to find out pH value of buffer Solution

pH = pKa + log  { [ salt ] / [ Acid ] }

= - log Ka + log { [ Salt ] / [ Acid ] }  

= - (1.8 x 10-5 ) + log { ( 0.1 ) / ( 0.1 ) }   

= - ( -4.7447 )

= 4.7447

Hence [ H+ ] = antilog of ( - 4.7447 )

= 5.55 x 10-5

Now when 10.0 ml of NaOH is added to 100ml of buffer solution , it will increase the [ OH- ] and decrease

[ H+ ] by equivalent concentration. , so Find out the concentration of OH-  in100ml of buffer solution using

relation N 1 V 1 = N 2 V 2   

N 1 x 100 = 0.100 x 10.0

N 1 = 0.01M

Thus the concentratin of hydroxyl ion in buffer solution would increase by 0.01 M , and accordingly the concentration of hydrogen ion would decrease by 0.01M

Therefore new [ H+ ] in buffer solution = ( 5.55 x 10-5 ) - 0.01  

= 0.009944M

& changed pH ( apply Handerson's equation again )

pH = pKa  + log (0.10 ) / (.009944)

= 4.7447 + 1.0021

= 5.7468

hence, Change in pH = 5.7468 - 4.7447

= 1.0021

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