What is the magnitude change in pH when 1.8g of NaOH is added 1L of a solution that is 0.100M in NH3 and 0.100 in NH4Cl? Kb for NH3 1.8 * 10^-5
pH of the original solution before NaOH is added is
pH = pKa + log (base)/(acid)
pH = 9.26 + log (0.1/0.1) = 9.26
millimols NH3 = 0.1 x 1000 = 100
mmols NH4Cl = 0.1 x 1000 = 100
mols NaOH added = 1.8/40 = 0.045 mols or 45 millimols.
.............NH4^+ + OH^- = NH3 + H2O
I............100......0......100
add..................45.............
C............-45....-45.......+45
E.............55......0........45
Plug the E line into the HH equation and solve for pH of the solution, then you want the difference between that and the original pH.
thank you
To determine the magnitude change in pH when NaOH is added to the solution, we need to calculate the moles of NaOH added and then analyze the effect of that amount on the pH of the solution.
First, let's calculate the moles of NaOH added to the solution. We are given the mass of NaOH (1.8g), and we can use its molar mass to convert grams to moles. The molar mass of NaOH is approximately 40 g/mol.
Moles of NaOH = Mass of NaOH / Molar mass of NaOH
Moles of NaOH = 1.8 g / 40 g/mol
Moles of NaOH = 0.045 mol
Now, let's consider the reaction that occurs when NaOH reacts with NH4Cl:
NaOH + NH4Cl → NaCl + NH3 + H2O
NaOH reacts with NH4Cl to form NaCl, NH3, and water (H2O). Here, NaOH acts as a strong base and NH4Cl acts as an acidic salt derived from NH4+ (ammonium ion) and Cl- (chloride ion).
As a result of this reaction, NaOH will neutralize a portion of NH4+ ions, shifting the equilibrium between NH3 and NH4Cl. This shift will affect the concentration of NH3 and NH4+ in the solution, ultimately influencing the pH.
To determine the magnitude change in pH, we need to calculate the initial and final concentrations of NH3 and NH4+ after the NaOH reaction.
Given:
Initial concentration of NH3 = 0.100 M
Initial concentration of NH4Cl = 0.100 M
Kb for NH3 = 1.8 × 10^-5
Since we are working with a 1 L solution, the initial moles of NH3 and NH4Cl can be calculated:
Initial moles of NH3 = Initial concentration of NH3 × Volume
Initial moles of NH3 = 0.100 M × 1 L
Initial moles of NH3 = 0.100 mol
Initial moles of NH4Cl = Initial concentration of NH4Cl × Volume
Initial moles of NH4Cl = 0.100 M × 1 L
Initial moles of NH4Cl = 0.100 mol
Now, let's analyze the reaction between NaOH and NH4Cl to determine how it affects the equilibrium between NH3 and its conjugate acid NH4+.
NaOH reacts with NH4Cl in a 1:1 ratio. Therefore, the number of moles of NH3 that will be neutralized by NaOH is equal to the moles of NaOH added, which we previously calculated as 0.045 mol. This means that 0.045 mol of NH3 will turn into its conjugate acid NH4+.
Final moles of NH3 = Initial moles of NH3 - Moles of NaOH
Final moles of NH3 = 0.100 mol - 0.045 mol
Final moles of NH3 = 0.055 mol
Final moles of NH4Cl = Initial moles of NH4Cl (same as initial)
Final moles of NH4Cl = 0.100 mol
To determine the final concentration of NH3 and NH4+, we divide the moles by the final volume (1 L):
Final concentration of NH3 = Final moles of NH3 / Volume
Final concentration of NH3 = 0.055 mol / 1 L
Final concentration of NH3 = 0.055 M
Final concentration of NH4+ = Final moles of NH4Cl / Volume
Final concentration of NH4+ = 0.100 mol / 1 L
Final concentration of NH4+ = 0.100 M
These concentrations will be used to calculate the new pH of the solution using the Kb expression for NH3.
Kb = [NH4+][OH-] / [NH3]
In the reaction of NH3 with water, OH- ions are produced. Since NaOH is a strong base, we can assume that it will dissociate completely, giving us an equal concentration of OH- ions.
Initial OH- concentration = Final OH- concentration = [OH-] = [NaOH]
[NaOH] = Moles of NaOH / Volume
[NaOH] = 0.045 mol / 1 L
[NaOH] = 0.045 M
Since the concentration of OH- ions is the same as the concentration of NaOH, we can substitute [OH-] with [NaOH] in the Kb expression.
Kb = [NH4+][NaOH] / [NH3]
Now, we can substitute the known values and solve for [NH3].
1.8 × 10^-5 = (0.100 M) × (0.045 M) / [NH3]
[NH3] = (0.100 M) × (0.045 M) / (1.8 × 10^-5)
[NH3] = 0.0025 M
Now that we have the final concentration of NH3, we can calculate its pOH and subsequently the pH.
pOH = -log10([NH3])
pOH = -log10(0.0025)
pOH ≈ 2.60
Since pH + pOH = 14 (at 25 degrees Celsius), we can calculate the pH.
pH = 14 - pOH
pH = 14 - 2.60
pH ≈ 11.40
Therefore, the magnitude change in pH when 1.8g of NaOH is added to the solution is approximately 11.40.