Quick Acids and Bases question please

[Nooro]

Hey I am kinda confused as to the explanation of a question from Kaplan's topical: Acids and Bases Test 1 question 12.

12. What is the approximate pH when .1 moles NaOH has been added to .2 moles hydrocyanic acid? (Ka of hydrocyanic acid given as 6 x 10^-10)
A. 10.0
B. 9.2
C. 7.0
D. 3.7

Correct answer B, 9.2

Explanation:
Hydrocyanic acid is a weak acid. When 0.1 moles NaOH has been added to 0.2 moles of the acid, half of the acid is converted into the salt sodium cyanide. This means that there are approximately equal amounts of the undissociated acid, HCN, and its conjugate base, the cyanide ion, CN−, from the salt. A solution containing approximately equal amounts of an acid and its conjugate base is a buffer solution. We can use the Henderson-Hasselbach equation, which you should memorize, to find the approximate pH for a buffer solution. The Henderson-Hasselbach equation says that pH of a buffer solution is approximately equal to the pKa of the acid plus the log of
ratio of the concentration of the conjugate base (in our case cyanide) to the acid. The pKa for hydrocyanic acid is 9.2. However, because half of the hydrocyanic acid has reacted with the sodium hydroxide, the amount of acid and conjugate base are both 0.1 mol and so the ratio of conjugate base to acid is 1. The log of 1 is zero and so the Henderson-Hasselbach equation simplifies to pH is approximately equal to pKa. The pH therefore is 9.2 and answer B is correct.

I Don't understand why the acid is .1 moles and its conjugate base are .1 moles. I'm sure this is a very stupid question and I just lack the fundamentals behind the Henderson-Hasselbach equation, so if someone could explain why its .1 and kinda just go over the equation with some examples I would be very grateful and it may help others =).

Thank you all for your time.

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[EndSong]

In this question you have a strong base and a weak acid. Strong bases always will deprotonate a weak acid as much as it possibly can - i.e. for every OH- the base produces it will snatch off a proton from the weak acid. Therefore, since there is a .1 M conc of OH and a .2 M conc of HCN, the .1 M of OH will take off .1 M of H+ from HCN giving you .1 M CN- and .1 M HCN. Hope this helps.

 

[Mutt]

That doesn't make sense because basicity is a function of [OH], not [CN-]....


Where is Q of Quimica when you need her? 😛

 

[dilated]

Basicity isn't intrinsically a function of [OH], it's just usually measured that way because of the behavior of aqueous solutions. I mean, if I have a syringe of t-butyl lithium, there's no OH- in there but it's still pretty damn basic.

When using Henderson-Hasselbalch the only thing you need to concern yourself with is the ratio of acid to base. OH/H3O will sort itself out.

 

[surgeonguy22]

Basicity doesn't have to be OH-

It could also be there is not alot of H+ in the solution....

 

[SpyPuts]

The quickest way to work this type of question is by considering the titration graph for the titration of a weak acid with a strong base.

The first thing you should know is that the best buffering region is produced when moles of strong base added is equal to 1/2 the original moles of weak acid (or vice versa). You're left with .1 moles of weak acid and .1 moles of its conjugate base. Think about H-H here. When concentration of weak acid is equal to the concentration of its conjugate base, pH = pKa + log(1) simplifies to pH = pKa.

Oh.. after typing that, I see you have that explanation. The only thing left to understand here is this reaction:

1 NaOH + 1 HCN ---> 1 H20 + 1 NaCN(aq)

0.1 moles of NaOH is added, so it is the limiting reactant. This neutralization reaction goes to completion because NaOH is a strong base, despite the fact that HCN is not normally completely dissociated. Now the concentration of HCN is reduced by 0.1 moles (or cut in half), and the concentration of free CN- ions is increased by 0.1 moles. HCN is approximately equal to CN-.