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Salt Hydrolysis Problems
- Write the Brønsted-Lowry reaction between the base CN!^ and the weak acid H 2 O.
CN!^ + H 2 O W HCN + OH!
- Write the Brønsted-Lowry reaction between the acid NH 4 +^ and the weak base H 2 O.
NH 3 +^ + H 2 O W NH 3 + H 3 O +
- Write the equilibrium expression for reaction 1 using a K (^) b.
Kb = [HCN] [OH!] [CN!]
- Write the equilibrium expression for reaction 2 using a K (^) a.
Ka = [NH 3 ] [H 3 O +] [NH 4 +]
In theory, you are able to work with these equilibria in a fashion similar to what you did in the previous exercise. However, if you attempt to look up the Kb and Ka which you need, you are not likely to find them. This is because the Ka for HCN and the Kb for NH 3 are listed and you can derive the respective K (^) b and Ka from them. This is because the product Ka Kb = Kw = 1.0 x 10!^14. Notice that this is true for the conjugate acid-base pair. For example:
Kb (CN!) = Kw/Ka (HCN)
Kb (CN!) = 1.0 x 10!^14 / 4.2 x 10!^10
Kb (CN!) = 2.4 x 10!^5
Also:
Ka (NH 4 +) = Kw/Ka (NH 3 )
Ka (NH 4 +) = 1.0 x 10!^14 / 1.8 x 10!^5
Ka (NH 4 +) = 5.6 x 10!^10
If you need to use these expressions, the problem is referred to as a salt hydrolysis.
- Calculate the pH of a solution which is 0.010 M in NaOCN. The Ka of HOCN is 3.5 x 10!^4.
Kb = Kw = [HOCN] [OH-^ ] Ka [OCN!]
1.0 x 10!^14 = 2.9 x 10!^11 = X^2 3.5 x 10!^4 0.
X = 5.3 x 10!^7 pOH = 6. Ans: pH = 7.
- Calculate the pH of a solution which is 0.10 M in NH 4 Cl. The Kb for NH 3 is 1.8 x 10!^5.
Ka = Kw = [NH 3 ] [H 3 O +] Kb [NH 4 +]
1.0 x 10!^14 = 5.6 x 10!^10 = X^2 1.8 x 10!^5 0.
X = 7.5 x 10!^6 pH = 5.
Ans: pH = 5.
- Calculate the pH of a solution which is 0.10 M in HOCN and 0.30 M in NaOCN. The Ka of HOCN is 3.5 x 10!^4. (Careful) Why? This is a buffer. Ka = [OCN!] [H 3 O +] [HOCN]
3.5 x 10!^4 = (0.30) X
X = 1.17 x 10!^4 pH = 3.
Ans: pH = 3.
- What is the pH of a solution formed by mixing 200.0 mL of 0.20 M NH 3 with 50.0 mL of water? Kb for NH 3 = 1.8 x 10!^5.
Note the dilution to 0.16 M NH 3
Kb = [NH 4 +] [OH!] [NH 3 ]
1.8 x 10!^10 = X^2
X = 1.7 x 10!^3 pOH = 2. Ans: pH = 11.
- What is the pH of a solution formed by mixing 200.0 mL of 0.2000 M NH 3 with 50.00 mL of 0.8000 M NH 4 Cl?
[NH 3 ] = 0.16 and [NH 4 +] = 0.16 a buffer.
pOH = 4.
Ans: pH = 9.
- What is the pH of a solution formed by mixing 200.0 mL of 0.2000 M NH 3 with 50.00 mL of 0.8000 M HCl?
Note to 4 sig. figs. this is neutralized ˆ is an Arrhenius salt and [NH 4 +] = 0.16 M
hydrolysis problem similar to 6
Ans: pH = 5.
- What is the pH of a solution formed by mixing 200.0 mL of 0.2000 M NH 3 with 50.00 mL of 0.8000 M HCl and 50.00 mL of 0.8000 M NH 4 Cl?
Note dilutions and addition of [NH 4 +]s = 0.133 + 0.
[NH 4 +] = 0.266 M
hydrolysis problem similar to 6 Ans: pH = 4.
Solubility and Dissociation Equilibria Problems Page 173
Solubility and Dissociation Equilibria Problems
- Calculate the calcium ion concentration for a solution in contact with CaF 2 if the fluoride concentration is 0.010 M and the Ksp for CaF 2 is 3.0 x 10!^11.
Ksp = [Ca2+][F^! ] 2 3.0 x 10!^11 = [Ca2+] (0.010) 2
Ans: 3.0 x 10!^7
- To a 0.1 M solution of Ca(NO 3 ) 2 is added enough Na 2 CO 3 to make the solution 0.10 M in CO 32!^ ion. If the Ksp for CaCO 3 is 4.8 x 10!^9 , what is the Ca2+ concentration?
Ksp = [Ca2+][CO 32^! ]
4.8 x 10!^9 = [Ca2+] (0.10)
Ans: 4.8 x 10!^8
- The Ka for the hydrogen carbonate ion (which is an acid) is 4.8 x 10!^11. Calculate the pH at which the Ca+^ in the above question will redissolve. (Hint: first calculate what the maximum CO 32!^ concentration must be for a 0.10 M Ca2+^ solution.)
Ksp = [Ca2+][CO 32^! ] 4.8 x 10!^9 = (0.10) [CO 32^! ] [CO 32^! ] = 4.8 x 10!^8 HCO 3!^ + H 2 O W CO 32!^ + H 3 O +^ Ka =
[CO 32^! ][H 3 O +] / [HCO 3^! ]
4.8 x 10!^11 = (4.8 x 10!^8 ) [H 3 O +] /(0.10) [H 3 O +] = 0.
Ans: 4.
- The Kd for CdCl 42!^ is 1.0 x 10!^4. What is the concentration of Cd2+^ in a solution of chloride which has 0.010 M of the CdCl 42!^ ion present. The Cl!^ concentration in this solution was measured and found to be 0.10 M.
CdCl 42!^ W Cd2+^ + 4Cl! Kd = [Cd2+][Cl^! ] 4 /[CdCl 42^! ] 1.0 x 10!^4 = [Cd2+] (0.10) 4 /(0.010)
Ans:
