Question

Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for any weak acid or weak base:
1. [ Select ] ["strong base", "weak base", "strong acid", "weak acid"] LiOH
2. [ Select ] ["weak acid", "strong acid", "strong base", "weak base"] HF
3. [ Select ] ["strong acid", "weak acid", "strong base", "weak base"] HCl
4. [ Select ] ["weak base", "strong base", "weak acid", "strong acid"] NH3
Ka expression: [ Select ] ["[H+][F-] / [HF]", "[Li+][OH-]/ [LiOH]", "[H+][Cl-} / [HCl]", "[NH4+] / [NH3]", "[HF] / [H+][F-}", "[LiOH] / [Li+][OH-]", "[HCl] / [H+][Cl-}", "none"]
Calculate the concentration of OHLaTeX: -? in a solution that has a concentration of H+ = 7 x 10LaTeX: -?6 M at 25°C. Multiply the answer you get by 1010 and enter that into the field to 2 decimal places.

Answer 1

Answer:

See explanation below

Explanation:

There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.

However we do not have a pH value here.

The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.

Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:

LiOH ---> Strong. If you try to dissociate :

LiOH ------> Li⁺ + OH⁻     The Li⁺ is a weak conjugate acid.

HF -----> Weak

HF --------> H⁺ + F⁻   The Fluorine is a relatively strong conjugate base.

HCl -----> Strong

This is actually one of the strongest acid.

NH₃ ------> Weak

Now writting the Ka and Kb expressions:

Ka = [H⁺] [F⁻] / [HF]

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Finally, to calculate the [OH⁻] we need to use the following expression:

Kw = [H⁻] [OH⁻]

Solving for [OH⁻] we have:

[OH⁻] = Kw / [H⁺]

Remember that the value of Kw is 1x10⁻¹⁴. So replacing:

[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶

[OH⁻] = 1.43x10⁻⁹ M

And now, multiplying by 10¹⁰ we have:

[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰

[OH⁻] = 14.29

Hope this helps