How many grams of NH4Cl do I have to add to 500 mL of 0.100 M NH4OH to have a pH of 9.40?
Kb= 1.8 x 10^-5
[OH-]= 2.51 x^-5 M
Kb = (NH4^+)(OH^-)/(NH4OH)
Solve for (NH4^+).
Then M NH4^+ = moles NH4^+/L
You know M and L, solve for moles.
Then moles NH4Cl = grams NH4Cl/molar mass MH4Cl. Solve for grams.
To find out how many grams of NH4Cl you need to add to 500 mL of 0.100 M NH4OH to achieve a pH of 9.40, we can use the concept of the Henderson-Hasselbalch equation and the relationship between pH, pOH, and the concentration of hydroxide ions.
Step 1: Calculate the pOH using the given [OH-] concentration.
pOH = -log10([OH-])
pOH = -log10(2.51 x 10^-5)
pOH ≈ 4.60
Step 2: Convert the pOH to pH using the relationship pH + pOH = 14.
pH + 4.60 = 14
pH ≈ 9.40
Step 3: Determine the concentration of NH4+ ions in the solution using the pOH value and the Kb value of NH4OH.
Kb = [NH4+][OH-] / [NH4OH]
1.8 x 10^-5 = [NH4+][2.51 x 10^-5] / 0.100
[NH4+] = (1.8 x 10^-5) * (0.100) / (2.51 x 10^-5)
[NH4+] ≈ 7.17 x 10^-5 M
Step 4: Calculate the moles of NH4+ ions in the 500 mL of NH4OH solution.
moles = concentration * volume
moles = 7.17 x 10^-5 M * 0.500 L
moles ≈ 3.59 x 10^-5 moles
Step 5: Convert the moles of NH4+ ions to grams using the molar mass of NH4Cl.
The molar mass of NH4Cl = 53.49 g/mol (14.01 g/mol for N + 1.01 g/mol for H + 35.45 g/mol for Cl).
grams = moles * molar mass
grams = 3.59 x 10^-5 moles * 53.49 g/mol
grams ≈ 1.92 x 10^-3 grams
Therefore, you need to add approximately 1.92 milligrams (mg) of NH4Cl to 500 mL of 0.100 M NH4OH to achieve a pH of 9.40.