General Chemistry Question

[Ibraiz]

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What will be the concentration of OH- at equilibrium if a mixture consisting 0.20 mole of NH4Cl and 0.30 mole of NaOH is diluted to 1 liter? (Assume ionization constant for ammonia, kb = 1.8 ^ -5)

NH3 + H2O ---> NH4+ + OH-

Achiever provided explanation but I am not understanding why they just subtracted .3 from .2 and got .1 as an answer without any calculation. Why is that?

 

[UndergradGuy7]

OH- is a strong base. NH4+ is an acid... I didn't think it was considered strong... but after some googling pKas on wikipedia says "The (pKa) of NH+4 is 9.25."

The pKa of HCl is 8 and HI is 10 and they are both strong acids, so NH4+ is a strong acid.

If you didn't have wikipedia you are given the Kb of NH3, in the reaction NH3 + H2O ---> NH4+ + OH- is 1.8 ^ 10^-5. The Kb is small and this means NH3 barely reacts. The opposite reaction should then give you a K a lot bigger than 1. Conjugate acid is strong if base is weak.

So strong acid and strong base react completely. 0.3-0.2 = 0.1 moles left.
0.1 moles / 1 L = 0.1 molar

My reasoning...

 

[DoCt0rPeTe]

Based on my knowledge, the pKa of HCl is negative. I don't know what source you were using, but it may very well be a -8.

The pKa of acetic acid is 4.75, so there is no way that HCl has a pKa of the same value.

The question is asking about a titration of Weak Acid with a Strong Base.
In order to fully titrate, the number of moles of base added must be equal to the number of moles of acid (NH4+) that you began with.

This occurs at 0.2 moles.

Now, you have added an EXTRA 0.1 moles of Base more than you need. This is termed "excess", and will thusly raise the pH.
How much excess do you have?
Roughly 0.3-0.2 moles = (0.1 mol Base excess) / (1 L )
gives the [OH-] concentration.


No disrespect to Undergrad because he ripped the DAT, but I'm pretty sure this is how this problem is solved.