Please show all work/steps. points will be awarded, I have a 100% rating.thanks
ID: 824315 • Letter: P
Question
Please show all work/steps. points will be awarded, I have a 100% rating.thanks
Part2:
7. A solution is 0.072M in both Mn(II) and Cu(II). It is desired to selectively precipitate each ion as a hydroxide salt in order to achieve a maximum separation. What pH will have this effect? (Ksp (Mn(OH)2 = 1.6 x 10-13; Ksp Cu(OH)2 = 4.8 x 10-20)
a) 5.32
b) 6.13
c) 7.87
d) 8.17
e) 9.42
8. Calculate the solubility of zinc oxalate in a 0.25 M zinc nitrate solution. Ksp ZnC2O4 = 2.7 x 10-8.
a) 1.2 x 10-4 M
b) 3.9 x 10-7 M
c) 1.1 x 10-7 M
d) 3.5 x 10-9 M
e) 1.6 x 10-11 M
9. It is desired to make 1.0 L of a buffer solution with a pH of 9.00. For this, hydrazine is chosen as the weak base in a concentration of 0.222 M (pKb=5.89). What must be the concentration of added hydrazinium iodide to achieve the desired pH?
a) 0.000172 M
b) 0.345 M
c) 0.15 M
d) 0.029 M
e) 1.71 M
10. Sometimes it is possible to alter the solubility of a compound by altering the pH of the solution in which it is dissolved. Cobalt(II) hydroxide has a Ksp of 1.3 x 10-15M. How would you expect the solubility of cobalt(II) hydroxide to change when the solution is made acidic?
a) solubility should rise as conjugate base reacts with H+.
b) solubility should rise as cobalt(II) forms a complex ion with added acid.
c) solubility should fall as acid drives solubility equilibrium to the left
d) solubility should be unaffected.
Part 3
1) What is the pH of a buffer solution that is 0.10 M Benzoic acid, HC7H5O2, and 0.15 M sodium benzoate
K benzoic acid = 6.6 x 10-5
ans.: pH = 4.4
2) A buffer is prepared by adding 45.0 ml of 0.15 M sodium Benzoate to 25.0 mL of 0.1 M Benzoic Acid. What is the pH of the final solution?
ans: pH = 4.6
3) What is the pH of a buffer solution that is prepared by making a solution 0.25 M NH3 and 0.15 M NH4+?
ans.: pH = 9.5
4) Calculate the pH of a buffer solution prepared by mixing 30 mL of 0.10 M HCl with 50 mL of 0.1 M NaCN?
ans.: pH = 9.2
5) Calculate the pH of a buffer obtained by mixing 500.0 mL of 0.10 M ammonia with 200.0 mL of 0.15 M HCl?
ans.: pH = 9.08
6) What is the pH of a buffer solution prepared by combining 250 mL of 0.3 M sodium dihydrogen phosphate to 150 mL of 0.22 M NaOH?
ans.: pH = 7.1
7) What initial concentration of aqeous sodium dihydrogen phosphate and aqeous sodium hydrogen phosphate is needed to prepare 500 mL a buffer with a pH of 6.89?
ans.: 250 mLs of 1.0 M NaH2PO4
and 250 mLs of 0.48 M Na2HPO4
8) How many grams of sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00? Ignore the volume change due to the addition of sodium acetate.
ans.: 29.52 grams of NaC2H3O2
9) Prepare 2 liters of a buffer solution having a pH = 7.5, using an appropriate acid and its corresponding conjugate base.
10) Prepare 1.5 liters of a buffer solution having a pH = 4.90, using NaOH and an appropriate acid, HX.
11) Prepare a buffer solution having a pH = 2.5, using 0.1 M HCl and an appropriate salt.
12) 1.5 mLs of 6 M HCl are added to 250 mLs of water (pH = 7 ). What is the resulting [H+] and pH?
13)A buffered solution of pH = 4.4 is 0.1 M Benzoic Acid, HC7H5O2, and 0.15 M Sodium Benzoate, NaC7H5O2. After addition of 1.5 mLs of 6.0 M HCl to 250 mLs of the buffered solution, what is the resulting pH?:
K benzoic acid = 6.6 x 10-5
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Explanation / Answer
7) C) 7.87
8) Given
ZnC204 ----> Zn+2 + C2042-
Ksp = ( Zn+2) ( C204)
Zn(N03)2 ---> Zn+2 + 2N03-
given Zn(N03)2 = Zn+2 = 0.25
2.7 x 10-8 = 0.25 x ( C2042-)
(c204) = 1.1 x 10-7
so the answer is option C c) 1.1 x 10-7 M
9) for a basic buffer
p0H = Pkb + log ( salt/base)
14- pH = Pkb + log (salt/acid )
14-9 = 5.89 + log ( salt /0.222)
(salt) = 0.029 M
so the answer is d) 0.029 M
10)
a) solubility should rise as conjugate base reacts with H+.
Part - 3
1) for acidic buffer
pH = pKa + log ( salt/acid )
pH = -log 6.6 x 10-5 + log ( 0.15/0.10)
pH = 4.36
2) pKa = 4.18
moles of benzoic acid = 25 x 0.1 /1000 = 2.5 x 10-3
moles of sodium benzoate = 45 x 0.15 /1000 = 6.75 x 10-3
pH = 4.18 + log ( 6.75 x 10-3 / 2.5 x 10-3 )
pH = 4.61
3) pKb = 4.74
for basic buffer
pOH = pkb + log (salt/acid)
poH = 4.74 +log ( 0.15/0.25 )
pOH = 4.5
pH = 14-4.5 = 9.5
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