Okay, this is a two-part question and the unit is acids and bases.
Ka for NH4 is 5.6 x 10^ -10
Kb for NH3 is 1.8 x 10^ -5
1. Determine the pH of a solution prepared by adding 0.0500 mole of solid ammonium chloride to 100mL of a 0.150 molar solution of ammonia.
2. If 0.0800 mole of solid magnesium chloride, MgCl2 is dissolved in the solution prepared in part (1) and the resulting solution is well-stirred, will a precipitate of Mg(OH)2 form? Show calculation to support your answer. (Assume the volume of the solution is unchanged. The solubility product constant for Mg(OH)2 is 1.5 x 10^ -11)
You have not asked a question.
1.
Use the Henderson-Hasslebalch equation to calcualte the pH of the solution.
2. From question 1, calculate (OH^-), then use that and (Mg^2+) to determine if (Mg^2+)(OH^-)^2 exceeds Ksp.
Post work if you get stuck.
I look up this date because when this was posted was when i was born.
1. To determine the pH of the solution, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa and the ratio of the concentrations of the conjugate acid and base forms. In this case, the conjugate acid is NH4+ and the conjugate base is NH3.
The Henderson-Hasselbalch equation is:
pH = pKa + log([A-]/[HA])
Given information:
Ka for NH4+ = 5.6 x 10^-10
Kb for NH3 = 1.8 x 10^-5
Since Ka and Kb are related by the autoionization of water (Kw = Ka x Kb), we can find Kw:
Kw = (5.6 x 10^-10)(1.8 x 10^-5) = 1.008 x 10^-14
The molar concentration of NH4+ in the solution can be calculated using the number of moles and volume:
[HA] = (0.0500 moles) / (0.100 L) = 0.500 M
The molar concentration of NH3 in the solution is given as 0.150 M.
Plugging in the values into the Henderson-Hasselbalch equation:
pH = log10(0.150/0.500) + log10((1.008 x 10^-14) / (5.6 x 10^-10))
pH = -0.477 + (-4.05)
pH = -4.53
Therefore, the pH of the solution is approximately 4.53.
2. To determine if a precipitate of Mg(OH)2 will form when 0.0800 moles of MgCl2 is dissolved in the solution prepared in part (1), we need to calculate the concentration of hydroxide ions (OH-) and the concentration of magnesium ions (Mg^2+).
From part (1), we know that there is a certain concentration of ammonia (NH3) and ammonium ions (NH4+). Ammonia can react with water to produce hydroxide ions:
NH3 + H2O → NH4+ + OH-
Since the volume of the solution is assumed to be unchanged, the moles of NH4+ formed from the reaction will equal the moles of OH- produced. Therefore, the concentration of OH- can be calculated from the concentration of NH4+.
Using the molar concentration of NH4+ calculated in part (1):
[NH4+] = 0.500 M
Since the reaction is 1:1, the concentration of OH- is also 0.500 M.
Now, we need to calculate the concentration of Mg^2+ in the solution. The dissociation of MgCl2 in water is as follows:
MgCl2 → Mg^2+ + 2Cl-
The moles of magnesium ions (Mg^2+) are equal to the moles of magnesium chloride (MgCl2):
[Mg^2+] = (0.0800 moles) / (0.100 L) = 0.800 M
Finally, we can use the solubility product constant (Ksp) to determine if a precipitate of Mg(OH)2 will form. The Ksp for Mg(OH)2 is given as 1.5 x 10^-11.
The expression for the solubility product constant is:
Ksp = [Mg^2+][OH-]^2
Plugging in the values:
Ksp = (0.800)(0.500)^2
Ksp = 0.100
Since the calculated Ksp (0.100) is greater than the given Ksp (1.5 x 10^-11), a precipitate of Mg(OH)2 will not form.
Therefore, no precipitate of Mg(OH)2 will form when 0.0800 moles of MgCl2 is dissolved in the solution prepared in part (1).