Answer:
Buffers would not result in precipitation are:
*1.0M HNO2 with 1.0M NaNO2 (Ka = 4.5x10⁻⁴; pKa = 3.347)
*5.6M C6H5OH with 1.0M KC6H5O (Ka = 1.3x10⁻¹⁰; pKa = 9.89)
*2.0M HCN with 1.0M NaCN (Ka = 4.0x10⁻¹⁰; pKa = 9.40)
Step-by-step explanation:
The Ksp of magnesium hydroxide, Mg(OH)₂, is:
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Ksp = [Mg²⁺] [OH⁻]² = 1.2x10⁻¹¹
The precipitation will occurs when the product [Mg²⁺] [OH⁻]² > Ksp. If [Mg²⁺] is 0.0173M:
[0.0173M] [OH⁻]² > 1.2x10⁻¹¹
[OH⁻]² > 6.936x10⁻¹⁰
[OH⁻] > 2.63x10⁻⁵M
pOH (-log [OH]) is 4.58, and pH (pH = 14-pOH) is:
pH = 9.42, if pH > 9.42, the precipitation will occurs
For the following options, using H-H equation, pH is:
*1.0M HNO2 with 1.0M NaNO2 (Ka = 4.5x10⁻⁴; pKa = 3.347)
pH = 3.347 + log (1.0M / 1.0M)
pH = 3.347. As pH < 9.42; the buffer solution will not result in precipitation.
*2.0M C6H5OH with 1.32M KC6H5O (Ka = 1.3x10⁻¹⁰; pKa = 9.89)
pH = 9.89 + log (1.32M / 2.0M)
pH = 9.71. As pH > 9.42; the buffer solution will result in precipitation.
*1.0M HCN with 2.0M NaCN (Ka = 4.0x10⁻¹⁰; pKa = 9.40)
pH = 9.40 + log (2.0M / 1.0M)
pH = 9.70. As pH > 9.42; the buffer solution will result in precipitation.
*5.6M C6H5OH with 1.0M KC6H5O (Ka = 1.3x10⁻¹⁰; pKa = 9.89)
pH = 9.89 + log (1.0M / 5.6M)
pH = 9.14. As pH < 9.42; the buffer solution will not result in precipitation.
*2.0M HCN with 1.0M NaCN (Ka = 4.0x10⁻¹⁰; pKa = 9.40)
pH = 9.40 + log (1.0M / 2.0M)
pH = 9.10. As pH < 9.42; the buffer solution will not result in precipitation.
*1.0M C6H5OH with 1.0M KC6H5O (Ka = 1.3x10⁻¹⁰; pKa = 9.89)
pH = 9.89 + log (1.0M / 1.0M)
pH = 9.89. As pH > 9.42; the buffer solution will result in precipitation.