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 Post subject: CHEM HELPPPP! Adding a strong acid to buffer?
PostPosted: Thu Apr 24, 2014 11:22 pm 
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A beaker with 105 ml of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100M . A student adds 6.60 ml of a 0.370 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased.


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 Post subject: CHEM HELPPPP! Adding a strong acid to buffer?
PostPosted: Fri Jun 22, 2018 9:56 am 
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5.00 = 4.760 + log acetate / acetic acid

5.00 - 4.760=0.24
10^0.24 = 1.74 = acetate / acetic acid

1.74 acetic acid = acetate

acetic acid + acetate = 0.100

acetia acid + 1.74 acetic acid = 0.100
acetic acid = 0.0365 M
acetate = 0.100 - 0.0365=0.0635 M

moles acetic acid = 0.105 L x 0.0365=0.00383
moles acetate = 0.0635 x 0.105 L=0.00667

moles H+ added = 6.60 x 10^-3 L x 0.370 =0.00244

H+ will react with acetate to give acetic acid

moles acetic acid = 0.00383 + 0.00244 = 0.00627
moles acetate = 0.00667 - 0.00244 =0.00423

pH = 4.760 + log 0.00423/ 0.00627=4.589
change = - 0.411



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