1. The pH of a solution prepared by dissolving 0.350 mol of acid in 1.00L of 1.1
ID: 1046875 • Letter: 1
Question
1. The pH of a solution prepared by dissolving 0.350 mol of acid in 1.00L of 1.10M of conjugate base is ________. The Kb for the conjugate base is 5.40 x 10^-4. (Assume the final volume is 1.00 L.)
a. 11.23
b. 1.66
c. 11.14
d. 2.77
e. none of these
2. Decreasing the pH of blood will cause hemoglobin to release ________.
a. H2
b. CO2
c. N2
d. Fe
e. O2
3. A 25.0-mL sample of 0.150 M hypochlorous acid is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base is added? The Ka of hypochlorous acid is 3.0 × 10-8.
a. 7.46
b. 7.58
c. 7.25
d. 4.43
e. 1.34
4. Suppose you have just added 100.0 ml of a solution containing 1.00 mole of acetic acid per liter to 500.0 ml of 0.100 M KOH. What is the final pH? The Ka of acetic acid is 1.77 × 10-5.
Explanation / Answer
1)
Answer
a) 11.23
Explanation
Ka = Kw/Kb = 1.00×10-14/5.40×10-4 = 1.85×10-11
pKa = - log(Ka) = - log(1.85×10-11)= 10.73
pH = pKa + log([Cojucate base/acid]
pH = 10.73 + log(1.10M/0.350M)
pH = 10.73 + 0.50
pH = 11.23
2)
b) CO2
3)
Answer
b) 7.58
Explanation
NaOH + HClO - - - - - > NaClO + H2O
Initial moles of HClO = (0.150mol/1000ml)×25ml = 0.00375
moles of NaOH added = (0.150mol/1000ml)×13.3ml = 0.001995
0.001995 moles of NaOH react with 0.001995 moles of HClO
to produce 0.001995moles of NaClO
After addition
No of moles of HClO = 0.00375 - 0.001995 =0.001755
No of moles of ClO- produced = 0.001995
Total volume = 25 + 13.3 = 38.3ml
[HClO] = (0.001755mol/38.3ml)×1000ml = 0.0458M
[ClO-]= (0.001995mol/38.3ml)×1000ml = 0.0521M
pKa = - log(Ka) = - log(3.0×10-8) = 7.52
pH = pKa + log([A-] /[HA])
pH = 7.52 + log(0.0521M/0.0458M)
pH = 7.52 + 0.06
pH = 7.58
4)
NaOH + CH3COOH - - - - - - - > CH3COONa + H2O
Initial moles of NaOH = (0.100 mol/1000ml)×500ml = 0.0500
moles of CH3COOH added = (1.00mol/1000ml)×100ml = 0.100mol
0.0500moles of NaOH react with 0.0500moles of CH3COOH to produce 0.0500moles of CH3COO-
after addition
No of moles of CH3COOH = 0.100- 0.050 = 0.050
No of moles of CH3COO- = 0.0500
Total volume = 600ml
[CH3COO-]= (0.0500mol/600ml)×1000ml = 0.0833M
[CH3COOH] = (0.0500mol/600ml)×1000ml = 0.0833M
pKa = - logKa= - log(1.77×10-5) = 4.75
Henderson-Hasselbalch equation is
pH = pKa + log([A-] /[HA])
pH = 4.75 + log(0.0833M/0.0833M)
pH = 4.75
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