How to Calculate pH and Choose the Best Indicator for Titration?

  • Thread starter Thread starter itsmybedtime
  • Start date Start date
  • Tags Tags
    Chemistry
Click For Summary
SUMMARY

This discussion focuses on calculating the pH of a solution formed by mixing 16.5 mL of 0.182 M HCN with 29.2 mL of 0.105 M NaCN, using the formula pH = pKa + log([A-]/[HA]). The correct pH calculation yields 9.32, highlighting the importance of understanding weak acid-base equilibria. Additionally, the best indicator for titrating 0.10 M CH3NH2 with 0.10 M HBr is determined by calculating the pKa of the conjugate acid CH3NH3+, which is essential for identifying the appropriate pH range for indicator selection.

PREREQUISITES
  • Understanding of weak acid-base equilibria
  • Familiarity with the Henderson-Hasselbalch equation
  • Knowledge of calculating pKa from Kb values
  • Proficiency in logarithmic and quadratic equations
NEXT STEPS
  • Learn how to calculate pKa from Kb for weak bases
  • Study the principles of acid-base titration and equivalence points
  • Research common indicators used in acid-base titrations
  • Practice problems involving weak acid and weak base equilibria
USEFUL FOR

Chemistry students, educators, and anyone involved in laboratory work related to acid-base titrations and pH calculations.

itsmybedtime
Messages
2
Reaction score
0

Homework Statement


1. a) Calculate pH of solution that results form mixing 16.5 mL of 0.182 M HCN (aq) with 29.2 mL of 0.105 M NaCN. Ka of HCN is 4.9*10^-10

b) and how could we determine what 'indicator' would be the 'best' to use for a titration between 0.10 M CH3NH2 with 0.10 M HBr?


Homework Equations


pH = Pka + log([A-]/[HA])
ka=kw/kb

The Attempt at a Solution


a) pH = -log(4.9*10^-10) + log(.105/.182), which turns out wrong, the answer is actually 9.32(why?).

b) found the kb of CH3NH2 to be 4.4*10^-4, thus ka is found from the given equation. anyway it gives something in the range of 10 or so, the answer is actually 4-6, why? These questions have no likes in the textbook that my professor issued and the material is confusing. Please help.
 
Physics news on Phys.org
itsmybedtime said:
log(.105/.182), which turns out wrong, the answer is actually 9.32(why?).

These are not concentrations in the final solution. You have not mixed 1:1 volumes (in which case at least ratio of the concentrations would be preserved).

b) found the kb of CH3NH2 to be 4.4*10^-4, thus ka is found from the given equation. anyway it gives something in the range of 10 or so, the answer is actually 4-6, why?

Hard to say not seeing what you did. However, you have a weak acid that - once dissolved - gives alkalic solution? Gimme a break :wink:

Equivalence point calculation.

Selecting indicators for acid-base titration.
 
itsmybedtime wrote:
b) found the kb of CH3NH2 to be 4.4*10^-4, thus ka is found from the given equation. anyway it gives something in the range of 10 or so, the answer is actually 4-6, why?

Borek responded:
Hard to say not seeing what you did. However, you have a weak acid that - once dissolved - gives alkalic solution? Gimme a break

Borek, we are not sure what he meant by "10 or so"; itsmybedtime is confused and maybe is not clear on what some calculated values (possibly miscalculated) mean.

itsmybedtime next wrote:
These questions have no likes in the textbook that my professor issued and the material is confusing. Please help.

If that is really true, then you really should check some alternative textbooks, either another General Chemistry book, or one or two analytical textbooks with the topic of Neutralization Titrations. Are you certain that your professor did not discuss the topic details of weak acid-base equilibria more thoroughly during class lectures? Some proressors do this to fill some inadequacies of the chosen course textbook.

Some advice about studying weak acids & bases: Understanding their equilibria often requires studying the topic more than once; learning can be much/somewhat better the second time, even a third time. Also, develop some skill with quadratic equations and logarithms. They are essential PRACTICAL arithmetic for weak acids and bases.
 
symbolipoint said:
Borek, we are not sure what he meant by "10 or so"; itsmybedtime is confused and maybe is not clear on what some calculated values (possibly miscalculated) mean.

I can be wrong, but my understanding was that itsmybedtime calculated pH to be 10. As we deal with an acid solution, this result is wrong at first sight - one should expect pH below 7. This is just a reflex, you don't need to check details of the calculations to know that they have to be wrong in this case. That's what I was aiming at.
 
itsmybedtime asked about:
b) found the kb of CH3NH2 to be 4.4*10^-4, thus ka is found from the given equation. anyway it gives something in the range of 10 or so, the answer is actually 4-6, why? These questions have no likes in the textbook that my professor issued and the material is confusing. Please help.

One possible source of trouble is that CH3NH2 is a BASE, NOT an acid. Did you make a mistake in calculating Kb ? Do you know that most of your Kb calculations for this will involve [OH-] and that you might be justified in neglecting hydronium ions?

In any case, if you are titrating a weak base using a strong acid titrant, and you know Kb for the base, can you predict about at what pH will be the equivalence point? This will give you some idea which indicator to choose.
 
symbolipoint said:
One possible source of trouble is that CH3NH2 is a BASE, NOT an acid.

That's the correct approach - calculate Ka of conjugate acid from known Kb, use this Ka to calculate pH.

At the equivalence point you have just a solution of CH3NH3+ - which is a weak, conjugate acid of CH3NH2. This acid determines pH of the solution.
 
thanks for the insights everyone! I will be working on problems like this all night tonight. :smile:
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
7K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
9K