A beaker with 170 mL of an acetic acid buffer with a pH of 5.000 is sitting on a

Question

A beaker with 170 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.70 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. And the answer is NOT 4.85!

Explanation / Answer

5.00 = 4.760 + log [CH3COO-] / [CH3COOH] 10^0.24 = 1.74 = [CH3COO-] / [CH3COOH] [CH3COOH] + [CH3COO-] = 0.100 [CH3COOH] = 0.100 - [CH3COO-] 1.74 = [CH3COO-] / 0.100 - [CH3COO-] 0.174 - 1.74 [CH3COO-] = [CH3COO-] [CH3COO-] = 0.0635 M [CH3COOH]= 0.100 - 0.0635 =0.0365 M moles acetate = 0.200 L x 0.0635 =0.0127 moles acetic acid = 0.200 x 0.0365 =0.00730 moles HCl = 0.00620 L x 0.260 M=0.00161 CH3COO- + HCl >> CH3COOH moles acetate = 0.0127 - 0.00161 =0.0111 moles acetic acid = 0.00730 + 0.00161 =0.00891 total volume =0.206 L cocnentration acetate = 0.0111 / 0.206 =0.0538 M concentration acetic acid = 0.00891 / 0.206 =0.0433 M pH = 4.760 + log 0.0538 / 0.0433 =4.85

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