In the lab, you prepare a buffer from a solution of chloroacetic acid (CCl3CO2H)
ID: 1038943 • Letter: I
Question
In the lab, you prepare a buffer from a solution of chloroacetic acid (CCl3CO2H) and chloroacetic acid sodium salt (CCl3CO2Na) by combining 1.25 mol chloroacetic acid sodium salt and 2.30 L of an aqueous 2.75 M chloroacetic acid solution. The pKa of chloroacetic acid is 2.85. Use this information to answer the following questions. (10) 2. What is the pH of your buffer solution? Show all work and label all calculations with appropriate units. (4) a. b. To your buffer solution from 2a, you add 3.50 mL 8.50 M NaOH What is the pH of this new buffer solution? Show all work and label all calculations with appropriate units. (3) To your buffer solution from 2b, you add 6.5 mL 10.5 M HCI. What is the pH of this new buffer solution? Show all work and label all calculations with appropriate units. c.Explanation / Answer
a)
pH of acidic buffer = pka + log(cCl3coo^-/CCl3COOH)
pka of CCl3COOH = 2.85
no of mol of CCl3COO- = 1.25 mol
No of mol of CCl3COOH = V*M = 2.3*2.75 = 6.325 mol
pH = 2.85 + log(1.25/6.325)
= 2.146
b)
pH of acidic buffer = pka + log((CCl3COO^- + NaOH)/(CCl3COOH-NaOH))
pka of CCl3COOH = 2.85
no of mol of CCl3COO- = 1.25 mol
No of mol of CCl3COOH = 6.325 mol
No of mol of NaOH added = M*V = 3.5*10^-3*8.5 = 0.03 mol
pH = 2.85 + log((1.25+0.03)/(6.325-0.03))
= 2.16
c)
pH of acidic buffer = pka + log((ch3coo^- + NaOH - HCl)/(CH3COOH-NaOH+HCl))
pka of CCl3COOH = 2.85
no of mol of CCl3COO- = 1.25 mol
No of mol of CCl3COOH = 6.325 mol
No of mol of NaOH added = M*V = 3.5*10^-3*8.5 = 0.03 mol
No of mol of HCl added = M*V = 6.5*10^-3*10.5 = 0.068 mol
pH = 2.85 + log((1.25+0.03-0.068)/(6.325-0.03+0.068))
pH = 2.13
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