27 (15 points) a. What is the pH of o.1uofo.1Macetic acid (CH,cooH)? Ka 1.8x105
ID: 930229 • Letter: 2
Question
27 (15 points) a. What is the pH of o.1uofo.1Macetic acid (CH,cooH)? Ka 1.8x105 a. What is the pH after you added 10mL of o.1M NaoH to o.1L of o.1M acetic acid? b. What is the pH after you added 10mLofo.1M HCI to o.1Lof o.1M acetic acid? c. What is the pH of the 0.1L buffer contains o.1M of sodium acetatelCH,cooNa) and o.1M of acetic acid d. What is the pH of previous buffer(question 4) after you added 10mLof o.1M NaoH? e. What is the pH of previous buffer(question 4) after you added 10mL of o.1M HCI?Explanation / Answer
27.
(a) CH3COOH <==> CH3COO- + H+
with x amount dissociated
Ka = 1.8 x 10^-5 = x^2/0.1
x = [H+] = 1.34 x 10^-3 M
pH = -log[H+] = 2.87
(b) moles of NaOH- = 0.1 M x 0.01 L = 0.001 mols
moles of CH3COOH = 0.1 M x 0.1 L = 0.01 mols
[CH3COONa] = 0.001 mols/0.11 L = 9.1 x 10^-3 M
[CH3COOH] remaining = 9 x 10^-3/0.11 = 0.082 M
pH = pKa + log([CH3COO-]/[CH3COOH])
= 4.745 + log(9.1 x 10^-3/0.082)
= 3.79
(c) [H+] = 0.1 x 0.01/0.11 = 9.1 x 10^-3 M
pH = -log(9.1 x 10^03) = 2.04
(d) pH of buffer
pH = 4.745 + log(0.1/0.1)
= 4.745
(e) mols of NaOH added = 0.1 M x 0.01 L = 0.001 mols
new [CH3COO-] = (0.1 x 0.1) + 0.001/0.11 = 0.1 M
new [CH3COOH] = (0.1 x 0.1) - 0.001/0.11 = 0.082 M
pH = 4.745 + log(0.1/0.082) = 4.83
(f) moles of HCl added = 0.1 M x 0.01 L = 0.001 mols
new [CH3COO-] = (0.1 x 0.1) - 0.001/0.11 = 0.082 M
new [CH3COOH] = (0.1 x 0.1) + 0.001/0.11 = 0.1 M
pH = 4.745 + log(0.082/0.1) = 4.66
new [CH3COOH] = 0.1 x 0.
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