Problem: What are the final hydrogen ion concentration and pH of a solution obtained by mixing 200 mL of 0.4 M aqueous NH₃ with 300 mL of 0.2 M HCL? (K_B = 1.8 X 10^-5)

 

My first attempt at solving:

@ t₀ I have 100% in the acid form (HCl)

 

mol HCl present = 0.3 L x 0.2 M = 0.06 mol HCl

 

mol HCl present = mol NH₃ required

0.06 mol HCl = (x amount of NH₃ in Liters) (0.4 M)

x L = 0.15 L = 150L of NH_3(aq) :larrow:that doesn't make sense to me...

 

My 2nd attempt at solving:

 

pOH = pK_b + log (mol_HCl / mol_NH3)

pOH = 4.74473 + log (.08/.06) = 4.86967

 

pH = 14 - pOH = 14 - 4.86967 = 9.13033

 

[H⁺] = 7.41E-10 M

 

The answer key states that the

[H⁺] = 1.66E-9 M and pH = 8.78

 

What am I doing wrong? Thank you in advance!!!

Basically, the formula (pOH = pK_b + log (mol_HCl / mol_NH3) is incorrect. The correct formula is pOH = pK_b+log([HB+]/), where HB+ is the protonated weak base (in this case, NH4+) and B is the weak base (NH3).

 

So, let's start from the beginning again. 300ml of 0.2M HCl is 0.06 moles of HCl. 200ml of 0.4M NH3 is 0.08 moles of NH3. 1 mole of HCl reacts with 1 mole of NH3, forming 1 mole of NH4Cl, so mixing 0.06 moles of HCl and 0.08 moles of NH3 gives you 0.06 moles of NH4Cl and a remaining 0.02 moles of NH3.

 

Then, using the correct formula (pOH = pK_b+log([NH4Cl]/[NH3]), the pOH is calculated to be approximately 5.22185, which equals to a pH of approximately 8.778, or rounded to 8.78, the answer in the answer key.

Edited October 27, 201213 yr by weiming1998