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?
See below
To solve these questions, we need to understand a few concepts related to acid-base chemistry and equilibrium calculations.
(a) To find the pKb for ammonia, we first need to know the Kb value. The Kb is the base dissociation constant and can be calculated using the equation:
Kb = [NH4+][OH-] / [NH3]
Given the Kb value for ammonia is 1.79 x 10^-5, we can rearrange the equation to solve for pKb:
pKb = -log(Kb)
Substituting the given Kb value into the equation, we can calculate the pKb.
(b) To find the pH of the solution described in the question, we need to determine if the resulting solution is acidic or basic.
The 0.100 M NH4Cl solution can act as a weak acid, producing NH4+ and Cl- ions. The 0.200 M NH3 solution acts as a weak base, producing NH3 and OH- ions.
The Cl- ions from NH4Cl do not affect the pH significantly, so we can focus on the NH3 and OH- ions. By adding NH4+ and NH3, we are creating a buffer solution.
To find the pH of the buffer solution, we need to calculate the concentration of OH- ions produced by the reaction between NH4+ and NH3. Since we have the starting concentrations and volumes, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
Where [A-] is the concentration of the conjugate base (NH3), and [HA] is the concentration of the weak acid (NH4+).
(c) To find the new pH of the solution after adding 0.200 grams of NaOH, we need to calculate the concentration of OH- ions produced by the NaOH in the given volume (assuming volume remains constant). Then, using the new concentrations of NH3 and NH4+ (from part (c)), we can calculate the new pH using the Henderson-Hasselbalch equation, similar to part (b).
(d) If equal molar quantities of NH3 and NH4+ were mixed in solution, we would have a 1:1 ratio of NH4+ to NH3. In this case, the solution would be a buffer solution with equal amounts of the weak acid and its conjugate base.
To find the pH of the solution, we can use the Henderson-Hasselbalch equation with equal concentrations of NH4+ and NH3.
Remember, in each case, we need to make sure to convert between moles, volumes, and concentrations using appropriate stoichiometry calculations if necessary.