Table v Part A A beaker with 1.60x102 mL of an acetic acid buffer with a pH of 5
ID: 1033198 • Letter: T
Question
Table v Part A A beaker with 1.60x102 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 7.20 mL of a 4.740. Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has 0.470 M HCl solution to the beaker. How much will the pHI change? The pK, of acetic acid is weak nd its decreased P View Available Hint(s) pair or ApH ey Submit X Incorrect; Try Again; 2 attempts remaining Provide Feedback Next > an EExplanation / Answer
ANSWER:
pH of buffer is given as:
pH = pKa + log[CH3COO-] / [CH3COOH]
5.0 = 4.74 + log[CH3COO-] / [CH3COOH]
log[CH3COO-] / [CH3COOH] = 0.26
[CH3COO-] / [CH3COOH] = 1.8
[CH3COO-] = 1.98 [CH3COOH] =====1
Further [CH3COO-] + [CH3COOH] = 0.1 =====2
Substituting equation 1 in 2
1.8[CH3COOH] + [CH3COOH] = 0.1
2.8 [CH3COOH] = 0.1
[CH3COOH] = 0.1 / 2.8 = 0.035M;
Therefore [CH3COO-] = 0.1 - 0.035 = 0.065M
Moles of [CH3COOH] = Molarity X volume = 0.035 X 160mL = 5.6 mmoles
Moles of [CH3COO-] = 0.065 X 160mL = 10.4 mmoles
Moles of HCl added = Molarity X Volume = 0.470 X 7.2mL = 3.384mmoles
3.384mmoles will react with 3.384mmoles of CH3COO- ion of the buffer as:
HCl + CH3COO- ------> CH3COOH + Cl-
Moles of CH3COO- left = 10.4 - 3.384 = 7.01 mmoles
Total Moles of CH3COOH after addition of HCl = 5. 6 + 3.384 = 8.98 moles
Molarity Of CH3COO- = mmoles / volume in mL = 7.01 mmol/ 167.2mL = 0.04 M
Total volume = 160mL (of buffer) + 7.2mL (of HCl) 167.2mL
Molarity Of CH3COOH = mmoles / volume in mL = 8.98 mmol/ 167.2mL = 0.05 M
pH = 4.74 + log[CH3COO-] / [CH3COOH] = 4.74 + log[0.04 / 0.05]
pH = 4.74 - 0.0969 = 4.6
Change in pH = 5.0 -4.6 = 0.4
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