# Calculating Ka and Kb values from estimated pH values

• TrueStar
In summary, the student is having trouble calculating estimated Ka and Kb values for solutions with ZnCl2 and NaCl. He is hoping that by the end of the week things will be clearer.
TrueStar

## Homework Statement

In our chemistry lab, we estimated the pH values of various solutions using six different indicators. Based on these estimated pH values, we must calculate [H+] and the estimated Ka or Kb values.

## Homework Equations

For pH - ph=-log[H+}
For Ka and Kb values - Concentrations of Product / Concentrations of Reactant

## The Attempt at a Solution

I think I'm having the most trouble with the salts, like NaCl and ZnCl2. My estimated pH values for NaCl and ZnCl2 are 6.7 and 4.3 respectively.

I think I'm supposed to set up an ICE table for problems likes these, but I'm not sure. With another solution we tried (NaCH3CO2), you clearly see that OH- is produced when added into water. In this case, we took out estimated pH (7.7) and subtracted that number from 14 to get 6.3 which is the pOH. From this, I calculated that the concentration of OH is about 5.01x10^-7. I divided the concentration of NaCH3CO2, which was 0.10, by this number and I think my estimated Kb value is about 5x10^-6.

I don't know how to translate these kinds of calculations to the salts. NaCl is neutral.

I do know that water goes through autoionization, but I'm not sure if this has any role in calculating the Ka/Kb values for NaCl and ZnCl2.

Long story short, I'm not sure how to calculate my estimated Ka/Kb values for NaCl and ZnCl2...even assuming I'm calculating them correctly for solutions like NaCH3CO2.

Thank you!

First step would be always to correctly identify the reaction responsible for pH change.

In the case of NaCl there is no reaction and there is no Ka nor Kb value.

In other cases you should start writing expressions for Ka (or Kb). Think what you can tell about concentrations of all substances involved. Perhaps you can use stoichiometry to calculate them?

Zn2+ seems to be acidic. Can you think of a reaction that will make it behave this way?

Last edited by a moderator:
Hi Borek:

Thank you for clarifying the issue with NaCl.

As for ZnCL2, it dissolves with water to product Zn2+ and 2Cl-. However, I think Zn2+ would be a highly charged metal. That means, I think, that when the salt dissolves in water, the Zn2+ will become hydrated.

Something like Zn2+ + H2O $$\rightarrow$$ Zn(H2O)2+ I'm not sure what the coefficients should be here. This also means that -

Zn(H2O)2+ + H2O $$\Leftrightarrow$$ Zn(H2O)(OH)+ + H3O+

That is what makes this acidic and this is the information I would use to set up my Ka equation. That would look like.

Ka = [Zn(H2O)(OH)+][H3O] / [Zn(H2O)2+][H2O]

I need to add H2O to this equation because it's an aqueous solution at this point. Then I set up my ICE table, but now I'm not sure what number to start with since I'm longer working with ZnCl anymore. Would I calculate the [H+] concentration and use it for H3O+ at equilibrium?

Thanks. Let me know if I've gone wrong anywhere in my post.

TrueStar said:
Zn(H2O)2+ + H2O $$\Leftrightarrow$$ Zn(H2O)(OH)+ + H3O+

That's the idea behind.

Ka = [Zn(H2O)(OH)+][H3O] / [Zn(H2O)2+][H2O]

You don't need water here. I know it sounds counterintuitive, but the way this is usually done we are assuming concentration of water is sonstant (and in fact in most solutions it doesn;t change much) and we make it a part of dissociation constant.

Then I set up my ICE table, but now I'm not sure what number to start with since I'm longer working with ZnCl anymore.

Concentration of hydrated Zn is identical to concentration of ZnCl2.

Last edited by a moderator:
Thank you Borek. :)

I'm discovering that these are topics that we haven't gone over in our lectures yet. I'm hoping by the end of this week, this will all be crystal clear.

At least I can get started on my lab report.

## What is the difference between Ka and Kb values?

Ka and Kb values represent the acid dissociation constant and base dissociation constant, respectively. They indicate the strength of an acid or base in a solution and are calculated from the equilibrium concentrations of the dissociated and undissociated species.

## How do I calculate Ka and Kb values from estimated pH values?

To calculate Ka and Kb values, you need the pH of the solution and the initial concentrations of the acid or base. For weak acids, Ka can be calculated using the Henderson-Hasselbalch equation, while Kb can be calculated for weak bases using the same equation with a slight modification. Alternatively, you can use the concentrations of the dissociated and undissociated species in the equilibrium expression to calculate Ka or Kb.

## Can Ka and Kb values be used to compare the strength of acids and bases?

Yes, Ka and Kb values can be used to compare the strength of acids and bases. The higher the Ka value, the stronger the acid, while the higher the Kb value, the stronger the base. This is because a higher Ka or Kb value indicates a greater extent of dissociation, which corresponds to a stronger acid or base, respectively.

## Is it possible to estimate Ka and Kb values without knowing the initial concentrations of the acid or base?

No, it is not possible to estimate Ka and Kb values without knowing the initial concentrations of the acid or base. These values are dependent on the concentrations of the dissociated and undissociated species in the solution, and without this information, the values cannot be accurately calculated.

## How can Ka and Kb values be used in real-world applications?

Ka and Kb values are important in many areas of science and industry. In chemistry, they can be used to determine the pH of a solution and predict the strength of acids and bases. In biochemistry, they are essential for understanding enzyme kinetics and protein structure. In environmental science, Ka and Kb values are used to monitor and regulate the pH levels of bodies of water. In industry, they are used in the production of pharmaceuticals, cleaning products, and food and beverage products.

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