I have a test on monday and I NEED to ace it in order to raise my grade > . <

ldsfkjaslkf
right now, acids and bases are killing me

can somebody help me with these problems?

A buffer solution is prepared by mixing the weak base ammonia (Kb=1.77x10^-5) with ammonium chloride. The concentration of NH3 is 0.10M and that of NH4Cl is 050M. What is the pH of this solution?

If a 0.00100 mole sample of NaOH is added to 1.0L of the solution, what is the new pH of the solution?

To a new 1.0L sample of the buffer, 0.010 mole of HCl is added. What does it react with?

What happens to the pH of the same solution if another 0.10 mole of HCl is added?

The secret to acid/base problems is to recognize what you have produced in the problem and determine the pH from that.

A buffer solution is prepared by mixing the weak base ammonia (Kb=1.77x10^-5) with ammonium chloride. The concentration of NH3 is 0.10M and that of NH4Cl is 050M. What is the pH of this solution?
Use the Henderson-Hasselbalch equation.
pH = pKa + log[(base)/(acid)]
base is NH3. acid is NH4Cl. pKa for NH3 is pKw-pKb.


If a 0.00100 mole sample of NaOH is added to 1.0L of the solution, what is the new pH of the solution?
NaOH is a strong base, it will react with the acid of the buffer.
NH4^+ + OH^- ==> NH3 + H2O
Therefore, the NH4^+ decreases due to the added NaOH and the NH3 increases. Substitute those new concns of acid/base into the HH equation and solve the for the new pH.


To a new 1.0L sample of the buffer, 0.010 mole of HCl is added. What does it react with?
Here the HCl will react with the base, NH3.
NH3 + HCl --> NH4Cl
You can see the the NH3 concn will be decreased while the NH4^+ will be increased. Substitute those into the HH equation and solve for the new pH.


What happens to the pH of the same solution if another 0.10 mole of HCl is added?
Just redo the last one.

To solve these problems, we need to understand the concept of buffers, weak bases, and the reaction between acids and bases.

1. pH of the buffer solution:
A buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. In this case, ammonia (NH3) acts as a weak base, and ammonium chloride (NH4Cl) acts as its conjugate acid.

To find the pH of the buffer solution, we need to determine the concentration of H+ ions. We can do this by comparing the concentrations of the weak base and its conjugate acid. Given that the concentration of NH3 is 0.10M and NH4Cl is 0.50M, we can set up an equation using the equilibrium constant Kb for ammonia:

Kb = [NH4+][OH-] / [NH3]

Since OH- concentration is determined by the concentration of NH4Cl, we can assume that [OH-] = [NH4+]. Let's call this concentration x.

Substituting the values into the equation, we get:

1.77x10^-5 = x^2 / 0.10

Solving for x, we find x = 1.33x10^-3. This is the concentration of OH- ions, which we can convert to the concentration of H+ ions by using the equation Kw = [H+][OH-] = 1x10^-14.

[H+] = 1x10^-14 / [OH-] = 1x10^-14 / 1.33x10^-3 = 7.52x10^-12

Using the pH formula, pH = -log[H+], we find pH ≈ 11.12.

Therefore, the pH of the buffer solution is approximately 11.12.

2. New pH after adding NaOH:
When NaOH is added to the solution, it reacts with the NH4+ ions to form water and ammonium hydroxide (NH4OH). Since NH4OH is a weak base, it will dissociate partially to release OH- ions.

To calculate the new pH, we need to determine the change in concentration of NH4+ ions and the change in OH- ions. Since we have a 0.00100 mole sample of NaOH, it will react with an equal number of NH4+ ions.

The concentration of NH4+ ions is initially 0.50M, and since 0.00100 mole of NaOH reacts, the new concentration of NH4+ ions will be 0.50M - 0.00100M = 0.499M.

To find the new OH- concentration, we divide the number of moles of NaOH by the total volume of the solution, which is 1.0L.

[OH-] = 0.00100M / 1.0L = 0.00100M

Now, we can calculate the new [H+] and thus the new pH using the equation Kw = [H+][OH-].

[H+] = 1x10^-14 / [OH-] = 1x10^-14 / 0.00100 = 1x10^-11.

Using the pH formula, pH = -log[H+], we find pH ≈ 11.00.

Therefore, the new pH of the solution after adding NaOH is approximately 11.00.

3. Reaction of HCl with the buffer solution:
When HCl is added to the buffer solution, it reacts with the weak base NH3 to form ammonium chloride (NH4Cl). This reaction consumes NH3 and increases the concentration of NH4+ ions.

4. Effect of adding more HCl to the solution:
Adding more HCl to the solution will further increase the concentration of NH4+ ions by reacting with the remaining NH3 in the buffer solution. This leads to an increase in the concentration of H+ ions. As the concentration of H+ ions increases, the pH of the solution will decrease and become more acidic.

Note: These calculations involve simplifying assumptions and may not fully account for all factors. It is essential to consult your textbook, professor, or reliable sources for a detailed understanding of the topic and any additional information.