Titration Question

[victorias]

Titrating 20 ml of 0.500M HCl with 10 ml of 0.500M of NaOH. What is the resulting pH?

My approach is:

We need to find the new concentration of the acid after the base has been added.

N1C1V1 = N2C2V2

(1)C1(20) = (1)(0.500)(10)

C1 = 0.25 M

pH = -log (0.25) = 0.6


In the answer, they are trying to determine how many moles of the acid are remaining

0.0050 Moles/0.03 L = 0.17 M

pH = -log (0.17) = 0.77


What is wrong with my approach?
When can we use this N1C1V1 = N2C2V2 equation? I know that this equation is not taking the total volume into account but what does the concentration or pH calculated from this represent?

Initial pH should be -log (0.5) = 0.3

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[NextStepTutor_3]

Hi there, great question! The N1C1V1 = N2C2V2 equation only works for complete neutralizations. In other words, it should be used at the equivalence point of a titration. To find out why, let's ignore normality for a minute (since we're dealing with NaOH and a monoprotic acid, both of which have normality values of 1). This gives us M1V1 = M2V2 (I used "M" to denote molarity, but C is fine as well).

Since molarity is measured in mol/L, M1V1 = M2V2 is saying the same thing as (mol / L)1 (L)1 = (mol / L)2 (L)2. This simplifies to moles 1 = moles 2 (or moles acid = moles base). Now we see why this equation can't be used all the time! During the titration of a monoprotic acid, the moles of original acid present is only equal to the moles of base added at the equivalence point. We certainly cannot use the equation at any old random time (which many students mistakenly do), since there are SO many situations in which moles of acid does NOT equal moles of base. (Be very careful here as well - many students confuse this equation with the Henderson-Hasselbalch, which deals with an acid and its own conjugate base. N1C1V1 = N2C2V2 deals with an acid and the base you're using to titrate it.)

Back to your question - you initially have 0.01 moles of HCl and 0.005 moles of NaOH. From this alone, we can see that moles of acid and moles of base are NOT equivalent! Rather than reaching the equivalence point, this reaction will involve OH- neutralizing only some of the HCl, leaving H+ present in solution. The way you originally did your math calculated the concentration of HCl that would have been neutralized by 10 mL of 0.500 M NaOH. But we don't have that amount of HCl - we have more, so some will be left over. For this reason, we need to do the math the way they did it in the answer key.

Hope this helps - let me know if you have any additional questions

 

[aldol16]

We need to find the new concentration of the acid after the base has been added.

N1C1V1 = N2C2V2

(1)C1(20) = (1)(0.500)(10)

C1 = 0.25 M

pH = -log (0.25) = 0.6

The problem with your approach is that you don't have 0.25 M left over. 0.25 M is a concentration and your volume has changed once you've added 10 mL of NaOH. So there's a new concentration that must be calculated. You added 10 mL of 0.500 M NaOH, which is equal to 0.005 moles. That will completely neutralize 0.005 moles of HCl. But you have 20 mL of 0.500 M HCl, which is 0.010 moles. So you will have 0.005 moles of HCl left, or 0.005 moles of protons. These protons will be dissolved in a total of 30 mL of solution, making the final proton concentration 0.005 moles/0.030 L = 0.17 M.

 

[victorias]

Thanks!

Just to clarify:

For initial pH values, I can use the initial concentration to calculate it or determine it from a titration plot if it is given.
At the equivalent point (is this also called end point?), I can use the n1c1v1 = n2c2v2 equation because this represents the complete neutralization point.
For any other point on the titration plot (eg. in cases where excess acid or excess base is left over), I would need to determine the new concentration using the moles of acid or base that is left over and taking the change in volume into account as well.

 

[NextStepTutor_3]

Yes, that's correct! Using the example you originally gave, let's say they asked for the initial pH. Since the original concentration of HCl (a strong acid) was 0.500 M, you could simply calculate -log (0.500) and get your answer.

For the sake of the MCAT, the endpoint can be considered the same as the equivalence point. Technically, the endpoint marks the point where an indicator changes color (for example, have you ever done a titration in lab using a phenolpthalein indicator? If titrating an acid with a base, the endpoint occurs when your indicator turns pink). But indicators are typically intended to change color as close to the equivalence point as possible, so you can consider the two basically synonymous. At the equivalence point, you can absolutely use your n1c1V1 = n2c2V2 - just be very careful with normality, especially if you're conducting a polyprotic titration.

At any other point, you should use stoichiometry to find out how much acid (or base) remains after the two species react, then calculate pH using that concentration. The one other case in which n1c1V1 = n2c2V2 can be very useful (if you're careful) is the half-equivalence point, as long as you adjust accordingly. For example, let's say that you are asked something regarding the half-equivalence point of a titration of HF with KOH. If you are given the molarity and volume of HF, along with the molarity of KOH, you can use n1c1V1 = n2c2V2 to calculate the volume of KOH required to reach the equivalence point. You can then cut this volume in half to find the volume required to reach the half-equivalence point. Of course, it's also good to remember that at the half-eq pt., pH = pKa - but that's another story. In short, you're understanding this properly - let me know if you have any other questions