Please explain and show work, Thank you. (Biochemistry) 1. What is the pH of a s
ID: 571730 • Letter: P
Question
Please explain and show work, Thank you. (Biochemistry)
1. What is the pH of a solution that results from adding 25 mL of 0.50 M NaOH to 75 mL of 0.50 M CH3CO2H? (Ka of CH3CO2H = 1.8 ´ 10-5)
____ 2. What is the pH of an aqueous solution of 0.30 M HF and 0.15 M F-? (Ka of HF = 7.2 ´ 10-4)
____ 3. What is the pH of an aqueous solution containing 0.50 M CO32- and 0.20 M HCO3-? (Kb of CO32- = 2.1 ´ 10-4)
____ 4. Which of the following combinations would be best to buffer an aqueous solution at a pH of 2.0?
a.
H3PO4 and H2PO4-, Ka1 = 7.5 ´ 10-3
b.
HNO2 and NO2-, Ka = 4.5 ´ 10-4
c.
CH3CO2H and CH3COO-, Ka = 1.8 ´ 10-5
d.
H2PO4- and HPO42-, Ka2 = 6.2 ´ 10-8
e.
NH4+ and NH3, Ka = 5.7 ´ 10-10
____ 5. Which of the following combinations would be the best to buffer an aqueous solution at a pH of 5.0?
a.
H3PO4 and H2PO4-, Ka1 = 7.5 ´ 10-3
b.
HNO2 and NO2-, Ka = 4.5 ´ 10-4
c.
CH3CO2H and CH3COO-, Ka = 1.8 ´ 10-5
d.
H2PO4- and HPO42-, Ka2 = 6.2 ´ 10-8
e.
NH4+ and NH3, Ka = 5.7 ´ 10-10
Explanation / Answer
1) No of mole of CH3COOH = (0.50mol/1000ml)*75ml =0.0375mol
No of mole of NaOH = (0.50mol/1000ml)*50ml = 0.0125mol
NaOH + CH3COOH - - - - - > CH3COONa + H2O
1:1 reaction
0.0125mole of NaOH react with 0.0125mole of CH3COOH to produce 0.0125 mole of CH3COONa
remaining mole of CH3COOH = 0.0375 - 0.0125 = 0.0250mol
No of mole of CH3COOH priduced = 0.0125mol
Total volume =100ml
[CH3COOH] =( 0.0250mol/100ml)*1000ml = 0.250M
[CH3COO-] = (0.0125mole/100ml)*1000ml = 0.125M
Henderson- Hasselbalch equation is
pH = pKa + log([A-] /[HA])
pKa of acetic acid = 4.75
therefore,
pH = 4.75 + log(0.125M/0.250M)
= 4.75 - 0.30
= 4.45
2) Ka of HF = 7.20*10-4
pKa = - logKa
= - log(7.20*10-4)
= 3.14
Henderson - Hasselbalch equation is
pH = pKa + log([A-] /[HA])
[A-] = [F-] = 0.15M
[HA] = [HF] = 0.30M
Applying the values
pH = 3.14 + log(0.15M/0.30M)
= 3.14 - 0.30
= 2.84
3) Kb of CO32- = 2.1*10-4
pKb of CO32- = 3.68
Henderson - Hasselbalch equation for base
pOH = pKb + log([BH+] / [B])
= 3.68 + log( 0.20M/0.50M)
= 3.68 - 0.40
= 3.28
pH + pOH = 14
pH = 14 - 3.28
= 10.72
4)
Answer
a) H3PO4 and H2PO4- , Ka1 = 7.5*10-3
Explanation
H3PO4 has the dissociation constant of 7.5*10-3
pKa of H3PO4 = 2.12
pKa of H3PO4 is near to our target pH 2, so this bufferer system is suitable
5)
Answer
c) CH3COOH and CH3COO-, Ka = 1.8*10-5
Explanation
Ka of Acetic acid = 1.8*10-5
pKa of Acetic acid = 4.75
pKa of acetic acid is near to the target pH, c is the answer