Kw = 1.0 x 10 - 14 at 25 °C 1. (3 pts) Suppose that you wanted to make a buffer
ID: 992304 • Letter: K
Question
Kw = 1.0 x 10-14 at 25 °C
1. (3 pts) Suppose that you wanted to make a buffer solution with a pH = 4.00. Which buffer below would be the "most effective" (as described in lecture)? Circle one answer.
(a) HCOOH/HCOO-, (Ka(HCOOH) = 1.8 x 10-4) (d) HOBr/OBr-, (Ka(HOBr) = 2.5 x 10-9) (b) HNO2/NO2-, (Ka(HNO2) = 4.5 x 10-4) (e) HF/F-, (Ka(HF) = 7.2 x 10-4)
(c) CH3COOH/CH3COO-, (Ka(CH3COOH) = 1.8 x 10-5)
2. Write the balanced neutralization reaction that occurs when KOH (aq) is added to a C6H5COOH/NaC6H5COO buffer. Write the equation in net ionic form.
3. Write the balanced neutralization reaction that occurs when HBr (aq) is added to a HC6H5NH2Cl/C6H5NH2 buffer. Write the equation in net ionic form.
4. A buffer is prepared by mixing 80.0 mL of 0.25 M NaOCl and 60.0 mL of 0.30 M HOCl at 25 °C. To the above solution is added 8.00 mL of 0.50 M KOH. Calculate the pH of the resultant solution at 25 °C. For HOCl, Ka = 3.5 x 10-8 at 25 °C. To receive full credit, all work must be explicitly shown, including the relevant neutralization and hydrolysis chemical reactions that are occurring, and do not use the Henderson- Hasselbalch equation.
Explanation / Answer
1) Among the given options, the buffer which is made by HCOOH/HCOO- will be the best option to make the buffer with pH 4.00.
2) The balanced neutralization equation is -
C6H5COOH (aq) + OH- (Aq) -------------> C6H5COO- (aq) + H2O (l)
3) C6H5NH2 (aq) + H+ (aq) <----------> C6H5NH3+ (aq)
4) The number of mols of HOCl = (0.30 M)(60 mL) = 18 mmol
The number of mols of NaOCl = (0.25 M)(80 mL) = 20 mmol
The number of mols of KOH added = (0.50 M)*8 mL = 4 mmol
Therefore, the number if mmol of HOCl consumed = 18 mmol-4 mmol = 14 mmol
Number of mmol of NaOCl consumed = 20 mmol + 4 mmol = 24 mmol
Therefore - pH = pKa + log [base] / [acid]
= 7.45+log (24/14) = 7.68
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