A 100 milliliter sample of 0.100-molar NH4Cl solution was added to 80 milliliters of a 0.200-molar solution of NH3. The value of Kb for ammonia is 1.79 x 10^-5.
(a) What is the value of pKb for ammonia?
(b) What is the pH of the solution described in the question?
(c) If 0.200 grams of NaOH were added to the solution, what would be the new pH of the solution? (assume that the volume of the solution does not change.)
(d) If equal molar quantities of NH3 and NH4+ were mixed in solution, what would be the pH of the solution?
what's the answer
I assume you can do (a).
(b) pH = pKa + log[(base)/(acid)]
Note this is pKa and NOT pKb.
(c)moles NaOH = grams/molar mass NaOH.
Exactly how much of this can you do? What do you not understand?
I don't understand C and D
The purpose of a buffer is to resist a change in pH. When an acid is added, one of the materials (the base--in this case NH3) neutralizes the acid forming MORE of the salt. So NH4^+ increases and NH3 decreases. When a base is added, it reacts with the acid (in this case NH4^+, thus decreasing NH4+ and increasing NH3.
can you help set up the equations i would need to use?
NH4^+ + OH^- ==> NH3 + H2O
Set up an ICE chart.
initial:
NH4^+ = mL x M = 100 x 0.1 = 10 millimoles
NH3 = mL x M = 80 x 0.20 = 16 millimoles.
change:
we add 0.2 g NaOH which is 0.2/40 = 0.005 moles or 5 millimoles.
NH3 = +5 millimoles
NH4^+ = -5 millimoles
equilibrium:
NH3 = 16 + 5 = 21 millimoles.
NH4^+ = 10 - 5 = 5 millimoles.
You may substitute millimoles in place of concn (since millimoles/mL = molarity and the mL (180 mL) appears in both numerator and denominator) OR you can divide millimoles/180 mL to arrive at concn for both base and acid and substitute those numbers. Plug those into the HH equation and solve for pH.
For part d, just set up the HH equation and pH = pKa + log (base/acid). The question is asking you to calculate pH if base and acid were equal. So plug in the same number (any number you choose) for base and acid and calculate. Note that the log of 1 = 0.
To answer these questions, we will need to apply several concepts from chemistry, including pKb, pH calculations, and the effect of additional chemicals on the pH of a solution. Let's break down each question step by step:
(a) To find the value of pKb for ammonia, we can use the formula: pKb = -log(Kb). Given that Kb for ammonia is 1.79 x 10^-5, we can calculate pKb as follows:
pKb = -log(1.79 x 10^-5)
Use a scientific calculator to calculate the negative logarithm and you'll get the value of pKb for ammonia.
(b) To determine the pH of the solution described in the question, we need to understand the relationship between pKb and pH in the context of the ionization of NH4Cl and NH3.
NH4Cl dissociates into NH4+ and Cl- ions, while NH3 acts as a base and accepts a proton, forming NH4+.
The equilibrium between NH4Cl and NH3 can be represented as follows:
NH4+ + H2O ⇌ NH3 + H3O+
Since the problem does not mention any other acidic or basic substances, we need to consider that the NH4+ ion will contribute to the overall acidity of the solution. Thus, we can assume the final solution is acidic.
To find the pH, we need to compare the initial concentrations of NH4Cl and NH3 and consider the Ka and Kb values of their respective ionizations. However, the problem statement does not provide the Ka value for NH4+, so we cannot directly calculate the pH. It's likely that additional information is required to answer this part.
(c) Adding 0.200 grams of NaOH to the solution will introduce OH- ions, which will react with H3O+ ions in the solution, forming water. This reaction is known as neutralization, and the new pH of the solution will depend on the concentration of OH- ions present and the remaining concentration of H3O+ ions.
To calculate the new pH, we need to account for the new concentration of OH- ions after the neutralization reaction. Since the volume of the solution doesn't change, we can calculate the concentration of OH- using the given weight of NaOH. From there, we can calculate the concentration of H3O+ and subsequently determine the new pH.
(d) If equal molar quantities of NH3 and NH4+ were mixed in solution, the resulting solution would be a buffer solution since it contains a weak acid (NH4+) and its conjugate base (NH3) in equal amounts. To determine the pH of a buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the base (NH3) and [HA] is the concentration of the acid (NH4+).
Given equal molar quantities, [A-] and [HA] will be the same, resulting in log(1) = 0. Therefore, the pH would be equal to the pKa value of NH4+.
Note: It's important to note that to answer these questions accurately, we may require additional information or specific values, so it's recommended to refer to the provided source or course materials for complete and precise calculations.