what is the pH of a solution prepared by adding 0.259g of ammonium chloride to 115mL of water? Kb of NH3 is 1.8x10^-5
mols NH4Cl = grams/molar mass = about 0.005 but you need to do that more accurately. Then 0.005/0.115 = about 0.05
........NH4^+ + H2O ==>H3O^+ + NH3
I.......0.05M...........0......0
C........-x.............x......x
E......0.05-x...........x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.05-x). Substitute from above and solve for x = (H3O^+) then convert to pH.
To determine the pH of a solution, we need to calculate the concentration of hydroxide ions (OH-) using the given Kb value. Here's how you can find the pH step-by-step:
Step 1: Convert the mass of ammonium chloride (NH4Cl) to moles.
Given mass of ammonium chloride = 0.259 g
Molar mass of NH4Cl = 53.49 g/mol
Number of moles = mass / molar mass
Number of moles of NH4Cl = 0.259 g / 53.49 g/mol
Step 2: Calculate the concentration of NH4Cl in the solution.
Given volume of water = 115 mL = 0.115 L
Concentration of NH4Cl = moles / volume
Step 3: Calculate the concentration of NH3 (ammonia) in the solution.
Since ammonium chloride (NH4Cl) dissolves in water, it will dissociate into NH4+ (ammonium) and Cl- (chloride) ions. For every one NH4+ ion formed, one NH3 molecule will also be formed.
Therefore, the concentration of NH3 is equal to the concentration of NH4+.
Step 4: Use the concentration of NH3 to calculate the concentration of OH- ions.
Kb (NH3) = [OH-] * [NH4+] / [NH3]
We know that [NH4+] = [NH3]
Kb = [OH-] * [NH4+] / [NH4+]
Step 5: Rearrange and solve for [OH-].
[OH-] = (Kb * [NH4+]) / [NH3]
Step 6: Calculate the concentration of NH4+.
Concentration of NH4+ = (moles of NH4Cl) / volume
Step 7: Calculate the concentration of NH3.
Concentration of NH3 = Concentration of NH4+
Step 8: Substitute the values into the equation for [OH-] to find the concentration.
[OH-] = (Kb * Concentration of NH4+) / Concentration of NH3
Step 9: Convert [OH-] to pOH using the formula pOH = -log([OH-])
pOH = -log([OH-])
Step 10: Calculate the pH using the formula pH = 14 - pOH
pH = 14 - pOH
By following these steps with the given values, you should be able to find the pH of the solution.
To find the pH of the solution, we need to determine the concentration of the resulting ammonium chloride (NH4Cl) solution and then calculate the concentration of hydroxide ions (OH-) using the given Kb value for ammonia (NH3).
Step 1: Calculate the number of moles of ammonium chloride (NH4Cl):
First, we need to determine the molar mass of NH4Cl, which is 14.01 g/mol for NH3, and 35.45 g/mol for Cl. So, NH4Cl has a molar mass of 14.01 + 35.45 = 49.46 g/mol.
To calculate the number of moles, divide the given mass (0.259 g) by the molar mass (49.46 g/mol):
Number of moles of NH4Cl = 0.259 g / 49.46 g/mol ≈ 0.00523 mol
Step 2: Calculate the concentration of the resulting ammonium chloride solution:
The volume of the solution is given as 115 mL. However, we need to convert this to liters since concentration is typically expressed in moles per liter (M):
Volume in liters = 115 mL * (1 L / 1000 mL) = 0.115 L
Concentration of NH4Cl = Number of moles / Volume in liters:
Concentration = 0.00523 mol / 0.115 L ≈ 0.0454 M
Step 3: Calculate the concentration of hydroxide ions (OH-):
The Kb value of NH3 is given as 1.8x10^-5. This value represents the equilibrium constant for the following reaction:
NH3 + H2O ⇌ NH4+ + OH-
Since NH3 is a weak base, we can assume that the concentration of NH3 is equal to the concentration of NH4+ (which is the same as the concentration of NH4Cl we calculated earlier). Therefore, the concentration of NH3 is 0.0454 M.
Using the Kb value, we can calculate the concentration of OH-:
Kb = [NH4+][OH-] / [NH3]
1.8x10^-5 = [0.0454][OH-] / 0.0454
[OH-] = 0.0454 * 1.8x10^-5 ≈ 8.2x10^-7 M
Step 4: Calculate the pOH of the solution:
The pOH is the negative logarithm (base 10) of the hydroxide ion concentration. Using the concentration we found in the previous step:
pOH = -log10(8.2x10^-7) ≈ 6.09
Step 5: Calculate the pH of the solution:
pH + pOH = 14 (at 25°C)
pH = 14 - pOH
pH = 14 - 6.09 ≈ 7.91
Therefore, the pH of the solution is approximately 7.91.