A beaker with 200 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop.?

A beaker with 200 mL of an acetic acid buffer with a pH of 5.00 is sitting on a bench top. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.70 mL of a 0.280M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.



Express your answer numerically to three decimal places. Use a minus (-) sign if the pH has decreased.



Ugh. I can't figure this dude out!A beaker with 200 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop.?
pH = pKa + log [acetate]/ [acetic acid]



5.00 = 4.760 + log [acetate]/ [acetic acid]



5.00 - 4.760=0.24



10^0.24 =1.74 = [acetate]/ [acetic acid]



[acetate] + [acetic acid ]= 0.100



[acetate] = 0.100 -[acetic acid]



1.74 = 0.100 - [acetic acid]/ [acetic acid]



1.74 [acetic acid] = 0.100 - [acetic acid]

2.74 [acetic acid]= 0.100

[acetic acid ]=0.0365 M

[acetate]= 0.100 - 0.0365 =0.0635 M



moles acetic acid = 0.0365 M x 0.200 L=0.00730

moles acetate = 0.0635 M x 0.200 L=0.0127



moles HCl added = 0.280 M x 0.00670 L=0.00188



CH3COO- + H+ = CH3COOH



moles acetic acid = 0.00730 + 0.00188 =0.00918

moles acetate = 0.0127 - 0.00188=0.0108

total volume = 200 + 6.70 = 206.70 mL =%26gt; 0.2067 L



[acetic acid ]= 0.00918/ 0.2067 =0.0444 M

[acetate]= 0.0108 / 0.2067 =0.0522



pH = 4.760 + log 0.0522/ 0.0444 =4.832



4.832 - 5.00 = - 0.168

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