1) Calculate the pH of a buffer solution which is 0.36 M in HNO 2 and 0.39 M in
ID: 1089282 • Letter: 1
Question
1) Calculate the pH of a buffer solution which is 0.36 M in HNO2 and 0.39 M in NaNO2.
Ka of HNO2 = 4.6x10-4 at 25oC
2) An acetic acid/sodium acetate buffer is made that is 0.45 M in acetic acid and 0.45 M in sodium acetate. Calculate the pH after 0.036 mol of KOH is added to 1.0 L of the buffer. (Assume no volume change.)
3) Which of the following buffer solutions will be the best buffer for a pH of 5.0?
0.25 M HCN and 0.25 M KCN
0.25 M HC2H3O2 and 0.25 M NaC2H3O2
0.25 M HClO2 and 0.25 M KClO2
0.25 M HOCl and 0.25 M NaOCl
a.0.25 M HCN and 0.25 M KCN
b.0.25 M HC2H3O2 and 0.25 M NaC2H3O2
c.0.25 M HClO2 and 0.25 M KClO2
d.0.25 M HOCl and 0.25 M NaOCl
Explanation / Answer
1)
Answer
pH = 3.37
Explanation
According to Henderson- Hasselbalch equation
pH = pKa + log([A-] /[HA])
[A-] = [NO2-] = 0.39M
[HA] = [HNO2] = 0.36M
Ka = 4.6×10-4
pKa = - log(Ka) = - log(4.6×10-4) = 3.34
substituting the values
pH = 3.34 + log(0.39M/0.36M)
= 3.34 + 0.03
= 3.37
2)
Answer
pH = 4.82
Explanation
No of mole of Acetic acid = 0.45mol
No of mole of sodium acetate = 0.45mol
No of mole of HCl added = 0.036mol
KOH react with Acetic acid
OH- + CH3COOH - - - - - > CH3COO- + H2O
stoichiometrically, 1mole of OH- react with 1mole of CH3COOH to produce 1mole of CH3COO-
0.036 mole of OH- react with 0.036 mole of CH3COOH to produce 0.036 mole of CH3COO-
After addition
No of mole of CH3COOH = 0.45 - 0.036 =0.414
No of mole of CH3COOH = 0.45 + 0.036 = 0.486
Volume = 1L
[CH3COOH] = 0.414M
[CH3COO-] = 0.486M
Applying Henderson- Hasselbalch equation
pH = 4.75 + log(0.486/0.414)
= 4.75 + 0.07
= 4.82
3)
Answer
0.25 HC2H3O2 and 0.25M NaC2H3O2
Explanation
pKa value of acetic acid is 4.75 which is close to the target pH 5.So, b is the answer.
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