A 50.0 ml sample of 0.0240 M NH(aq) is titrated with aqueous hydrochloric acid. What is the pH after the addition of 15.0 mL of 0.0600M HCl(aq)? (Kb of NH3= 1.8 x 10^-5)
Initial amount of NH3 (0.0240)(0.05) = .0012 mol
Amount of HCl added (0.0600)(.015) = .009 mol
Amount of NH3 after reaction = .0012 - .0009 =.003 mol
Amount of NH4^+ after reaction = .0009
pOH = pKb + log (concentration of BH+/concentration of B)
pOH = 1.8 X 10^-5 + log [.0009]/[.0003]
1.85x10^-5 + log 3
pOH = 1.85x10^-5 + .47712 = .47713
Something is wrong. I think that I have a problem with the amount or concentration of NH4^+ after reaction.
I think you have two problems but neither is earth shaking.
First, you wrote pKb but typed in and calculated with Kb, not pKb.
pKb is -log Kb = 4.74 according to my figures.
Then pOH = pKb + log acid/base
I get 5.22 for pOH.
Second, the problem asks for pH so you need to subtract this from 14 to get 8.77. I think, however, that you have done it the hard way.
I find trying to remember two different formulas, one for pH and one for pOH is difficult. So I remember only one.
pH = pKa + log base/acid.
pH = 9.25 + log (0.0003/0.0009) = 8.77 directly without the subtraction step. Check my work.
Amount of ammonia initially = 0.050 x 0.0240 =0.0012 mol.
Amount of HCl added =0.015 x 0.0600 = 0.0009 mol.
Amount of ammonia left = 0.0012-0.0009 =0.0003 mol.
Amount of NH4+ ion = Amount of HCl added = 0.0009 mol.
pOH= pKb + log(NH4+/NH3)
pOH= -log(1.8 x 10-5) + log(0.0009/0.0003)
pOH= 4.74 + 0.477 =5.22.
pH =14-pOH.
pH = 14-5.22= 8.78.
Hence, option A) is correct.
To find the pH after the addition of 15.0 mL of 0.0600 M HCl(aq) to the 50.0 mL sample of 0.0240 M NH3(aq), you need to consider the reaction that occurs between NH3 and HCl.
The balanced equation for the reaction is:
NH3(aq) + HCl(aq) → NH4+(aq) + Cl-(aq)
From the given information, you correctly calculated the initial amount of NH3 as 0.0012 mol. However, the amount of NH4+ produced is equal to the amount of HCl added because the reaction is 1:1.
So, the amount of NH4+ after the reaction is 0.009 mol (the same as the amount of HCl added).
Now, let's recalculate the pOH and pH taking into account the correct amount of NH4+:
Amount of NH4+ after reaction = 0.009 mol
Amount of NH3 after reaction = 0.0012 mol - 0.009 mol = -0.0078 mol
Here, we notice that the amount of NH3 after the reaction is negative, which indicates that it is completely consumed in the reaction. In this case, we have an excess of acid (HCl) that is not neutralized, resulting in a strong acidic solution.
Since the concentration of NH3 after the reaction is 0 mol/L, the concentration of NH4+ is 0.009 mol / 0.065 L (total volume of the solution) = 0.1385 M.
Now, let's recalculate the pOH:
pOH = pKb + log([NH4+]/[NH3])
pOH = -log(1.8 x 10^-5) + log(0.1385/0)
pOH = -log(1.8 x 10^-5)
Using a calculator, we find that pOH ≈ -4.74
To find the pH, we can use the equation pH + pOH = 14:
pH + (-4.74) = 14
pH ≈ 14 + 4.74
pH ≈ 18.74
Therefore, the pH after the addition of 15.0 mL of 0.0600 M HCl(aq) to the 50.0 mL sample of 0.0240 M NH3(aq) is approximately 18.74.