Aqueous equilibria, buffered solutions

In summary, the conversation involves two problems in chemistry, one involving the calculation of moles of Na2CO3 needed to achieve a pH of 10,00 and the other involving the equilibrium constant for the reaction of Cr(OH)3 and OH- to form Cr(OH)4-. The problem is solved using the Henderson-Hasselbalch equation and the knowledge of equilibrium constants.
  • #1
silenzer
54
0
I have two problems.

Homework Statement



How many moles of Na2CO3 must be dissolved in 250 mL of 0,125 M NaHCO3 so that the solution has a pH of 10,00?


Homework Equations



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The powers of the concentrations are the coefficients in the chemical equation. The products of the equation are in the numerator, and the reactants in the denominator.

The Attempt at a Solution



I have an idea for a solution, but I have no way of checking whether it's correct. My solution is as follows.

The sodium ions can be ignored because they have no significant effect on the pH.

The equation is: HCO3- ⇔ H+ + CO32-

I set [CO3]2-=y, and to equilibrium this concentration changes to y+x, but I assume that x is very small compared to y so y+x ≈ y. The same applies to HCO32-. The initial concentration of that is 0,5 M, and it occurs change towards equilibrium that is 0,5 M - x, which is essentially 0,5 M.

If I'm allowed to do the above, the problem is easy. I know Ka for HCO32- so all I have to do is plug everything I know into the Ka equation and solve for [H+].

Homework Statement



In a very basic solution the insoluble salt Cr(OH)3 (Ksp=1,6*10-30) dissolves and forms the metal complex Cr(OH)4- (Kf=8*1028). Calculate the equilibrium constant, K, for the reaction:
Cr(OH)3(s) + OH-(aq) ⇔ Cr(OH)4-(aq).

Homework Equations



Same as in the first problem.

The Attempt at a Solution



I'm not sure how to approach this exact problem. I know how to solve simpler problems that are showcased in my textbook, but there you're given numbers like molarities to work with. Here I only have equilibrium constants.
 
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  • #2
  • #3
Thanks for the reply.

1. Yes I have, but to my knowledge its only use is that you calculate pH in one step instead of two. I'm wondering whether my y+x=y and 0.5-x=0.5 assumptions are justified.

2. Thanks, I'll try that.
 
  • #4
Well, it is possible to solve buffer problems using ICE tables. Whether your assumptions are correct you would have to check, solve the problem using them and see if your x and y have values that fit the approximations.

Just note that

silenzer said:
If I'm allowed to do the above, the problem is easy. I know Ka for HCO32- so all I have to do is plug everything I know into the Ka equation and solve for [H+].

concentration of H+ is not what you are looking for. Besides, initial concentration of HCO3- is not 0.5M. Are you sure you are solving the problem you have listed?
 
  • #5
It's this problem, I just made stupid mistakes. You're correct of course, it's 0.125 M.

Oh, silly me. Of course I'm not supposed to find H+. Well, then I don't know what to do. Now I have a y which is my unknown, but I also know nothing abut the concentration of H+. Do you have any ideas?

edit: Woops, of course I know [H+]. The pH is supposed to be 10,00 at equilibrium. The case is closed! Thanks for the help.
 
Last edited:
  • #6
My ideas? Obvious: use the HH equation.
 

FAQ: Aqueous equilibria, buffered solutions

What is an aqueous equilibrium?

An aqueous equilibrium is a state in which the concentrations of reactants and products in a solution remain constant over time. This occurs when the rate of the forward reaction is equal to the rate of the reverse reaction.

What is a buffered solution?

A buffered solution is a solution with the ability to resist changes in pH when small amounts of acid or base are added. This is due to the presence of a weak acid and its conjugate base, or a weak base and its conjugate acid, which can neutralize any added acid or base.

How do you calculate the pH of a buffered solution?

The pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the acid dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

How do you prepare a buffered solution?

A buffered solution can be prepared by combining a weak acid or weak base with its conjugate base or conjugate acid in specific ratios. For example, to prepare a buffered solution with a pH of 7, you could mix equal amounts of acetic acid (weak acid) and sodium acetate (conjugate base) in water.

How does temperature affect aqueous equilibria?

Increasing the temperature can shift the equilibrium of an aqueous solution towards the endothermic reaction, and decreasing the temperature can shift it towards the exothermic reaction. This is due to the effect of temperature on the equilibrium constant (K), which is a measure of how much reactants are converted to products at equilibrium.

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