[viperjo]

So heres my deal, i've got a titration problem with a monoprotic weak acid and a strong base and I need to find the half equivalence point and equivalence point.  The problem also asks to find the initial pH before titration and pH after some of the strong base has been added.  My problem comes in at the equivalence point.

The problem gives me 50mL of 0.25M HCN (Ka of 6.2x10^-10) and 0.15M NaOH.  Finding the initial pH and pH after 25mL being added posed little problem.  When asked to find the half and equivalence point I go completely blank.  I have no graph to use a geometric method of finding the half and full E-point, otherwise this would have taken all of about 3 minutes.  My gut tells me I should use M1V1 to find the volume and molarity at which my A-=HA but that process has yielded a pH less than when the 25mL of NaOH had been titrated.

My work is as follows.

50mL(0.25M HCN) = .15M(xV NaOH)
83.3mL = xV NaOH

50mL(0.25M HCN) = xM NaOH (133.3mL solution)
xM NaOH = .0938
83.3mL (0.15M NaOH) = xM HCN (133.3mL solution)
xMHCN = .0938

Using Henderson-Hasslebach  means pH = pKa + log[A-]/[HA]
pH = 4.9 + log(.0938/.0938)
pH = 4.9

This makes sense by the definition of the half equivalence point since pH = pKa at that point.  The calculations I did above should be for the equivalence point though correct?  It also confuses me because I am titrating the strong base into the weak acid, if I am understanding the problem correctly, therefore the pH should be going UP instead of going DOWN.  When I solved for the pH after adding 25mL of the NaOH I came up with 5.42.

If anyone can point me in the right direction I would be very grateful.

[Borek]

Honestly, I have no idea what you did and why. 0.0938 is a final concentration of NaCN - why do you put it into HH equation? Both in nominator and denominator?

First things first - what is pH before NaOH is added?

Where did you get Ka=4.9 from?

[viperjo]

You and I are in the same boat then.  I calculated the .0938 to try and find the 1/2 equivalence point.   I made the assumption that CN- ion is my A- in the HH equation and that the M1V1 equation also told me the [HA] concentration.

The pH before adding NaOH came out to be 4.9.  Taking the Ka of HCN (6.2x10^-10) and multiplying by the given molarity to get 1.55x10^-10.  I used a rice/reaction table to find x^2, set that equal to the 1.55x10^-10, solved for x, and finally took the negative logarithm to arrive at my pH.

Ka= x^2/[HA]
6.2x10^-10 = x^2/(.25)
I use x^2 to = [H3O+] and [A-], both of which are unknown/non-existent at initial concentrations.
1.55x10^-10 = x^2
1.22x10^-5 = x  This in turn equals my [H3O]initial
pH = -log(1.22x10^-5) = 4.9

I am confident that at least this much of the problem is correct.  Finding the pH after the 25mL of NaOH is added seems to be a sort of check on your work to make sure you are going in the right direction.  Somehow I managed to mess up still lol.

[Borek]

You and I are in the same boat then.

Not exactly - I know how to find the answer

I calculated the .0938 to try and find the 1/2 equivalence point.   I made the assumption that CN- ion is my A- in the HH equation and that the M1V1 equation also told me the [HA] concentration.

Are you sure you have not forget there is a neutralization reaction going on? Besides, you have Ka value, no need for any calculations.

The pH before adding NaOH came out to be 4.9.  Taking the Ka of HCN (6.2x10^-10) and multiplying by the given molarity to get 1.55x10^-10.  I used a rice/reaction table to find x^2, set that equal to the 1.55x10^-10, solved for x, and finally took the negative logarithm to arrive at my pH.

Ka= x^2/[HA]
6.2x10^-10 = x^2/(.25)
I use x^2 to = [H3O+] and [A-], both of which are unknown/non-existent at initial concentrations.
1.55x10^-10 = x^2
1.22x10^-5 = x  This in turn equals my [H3O]initial
pH = -log(1.22x10^-5) = 4.9

Final result OK, you are aware of the fact you have used an approximation for [HCN] concentration?

So, you have correct starting point. You have posted somwehere correct information about midpoint pH - although you have then dismised this information and started some calculations which I was not able to follow. What is the midpoint pH again? How is it related to pKa?

[viperjo]

Final result OK, you are aware of the fact you have used an approximation for [HCN] concentration?

So, you have correct starting point. You have posted somwehere correct information about midpoint pH - although you have then dismised this information and started some calculations which I was not able to follow. What is the midpoint pH again? How is it related to pKa?

I am aware that my [HCN] is approximate.  Midpoint pH is equal to pKa at 1/2 equivalence point.  That said, and now fresh in my mind, would I simply multiply my pKa by 2 for the equivalence point?  If so, then my 1/2 E-point is equal to 9.2 (-log(6.2x10^-10) and then my full E-point is 8.91 (-log(1.24x10^-9)).  This doesn't quite make sense though.  pH should be going UP rather than DOWN.  So that leads me to think that I should divide by 2 and end up with 9.5 as my equivalence point pH (-log3.1x10^-10).

If thats all then I feel like one dumb duck for having tried to over think it all.

[Borek]

Midpoint pH is equal to pKa at 1/2 equivalence point.

If so, then my 1/2 E-point is equal to 9.2 (-log(6.2x10^-10)

Bingo!

As for endpoint - you have a solution of CN^- (concentration easy to calculate). This is a weak base.

pKb & C -> pOH -> pH

Follow the path, don't turn around, you'll get there!