A beaker with 100 mL of an acetic acid buffer with a pH of 5.00 is sitting on a

Question

A beaker with 100 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 0.100 M . A student adds 6.80 mL of a 0.460 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Explanation / Answer

below procedure is correct. please change constants to get correct result. ThanQ let x = Molarity acid and let y = molarity acetate x + y = 0.1 5.00 = 4.74 + log y / x 5.00 = 4.74 + log 0.1 - x / x 10^0.26 = 1.82 = 0.1 - x / x 1.82 x = 0.1 - x 2.82 x = 0.1 x = 0.0355 M y = 0.1 - 0.0355 = 0.0645 M moles acid = 0.0355 x 0.165 L = 0.00586 moles acetate = 0.0645 x 0.165 = 0.0106 Moles HCl = 0.400 x 0.0044 L = 0.00176 CH3COO- + H+ >> CH3COOH moles acid = 0.00586 + 0.00176 = 0.00762 moles acetate = 0.0106 - 0.00176 = 0.00884 total volume = 165 + 4.4 = 169.4 mL = 0.169 L concentration acid = 0.00762 / 0.169 = 0.0450 M concentration acetate = 0.00884 / 0.169 = 0.0523 M pH = 4.74 + log 0.0523 / 0.0450 = 4.81

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