Calculate The PH Of A Solution That Is 0.181 M NH3 And 0.24 M NH4Cl.

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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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This 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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