Calculate The PH Of A Solution That Is 0.181 M NH3 And 0.24 M NH4Cl.
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Search Find an Online Tutor Now Ask Ask a Question For Free Login Chemistry Calculate the pH of a solution that is 0.181 M NH3 and 0.24 M NH4Cl. Calculate the pH of a solution that is 0.181 M NH3 and 0.24 M NH4Cl.
Kb of NH3 is 1.8 x 10-5.
The answer is 9.13
please explain the steps, thank you :)
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Best Newest Oldest By: Best Newest OldestThis is a buffer since you have a weak base (NH3) and the conjugate acid (salt) which is NH4Cl, or simply NH4+. For such calculations you can use the Henderson Hasselbalch equation.
For a base, the HH equation is pOH = pKb + log [salt]/[base]
For an acid, the HH equation is pH = pKa + log [salt]/[acid]
Since we are dealing with a weak base in the current problem, we will use the former:
pOH = pKb + log [NH4Cl]/[NH3]
To find the pKb from the Kb, simply take the negative log of the Kb. -log1.8x10-5 = 4.745
pOH = 4.745 + log [0.24]/[0.181]
pOH = 4.745 + log 1.326
pOH = 4.745 + 0.123 = 4.868
Then to find the pH, we can use pH + pOH = 14 and pH = 14 - pOH
pH = 14.00 - 4.868
pH = 9.13
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