Weak Acid-Base Equilibria: Videos & Practice Problems

To determine the original molarity of a weak acid with a given acid ionization constant (Ka) of 4.7 × 10^-3 and a pH of 4.12, first calculate the hydronium ion concentration using the formula ${10}^{-}$, resulting in 7.59 × 10^-5 M. Set up the equilibrium expression $K=\frac{\mathrm{[A^-]}}{\mathrm{[H₃O⁺]}}$ and solve for the formal concentration, yielding 7.71 × 10^-5 M.

Weak Acid-Base Equilibria Calculations 1

It asks us what is the original molarity of a solution of weak acid with a \( K_a \) of \( 4.7 \times 10^{-3} \) and a pH of 4.12 at 25 degrees Celsius. Now, the key here is realizing that we're dealing with a weak acid. Remember, we know it's a weak acid because its \( K_a \) value is less than 1. With a weak acid, we're going to say that our generic weak acid, we can say represents weak acids, weak bases react with water in order to establish the equilibrium. Remember, the acid here according to the Bronsted-Lowry definition will donate an \( H^+ \).

It's going to become \( A^- + H_3O^+ \). We have initial change equilibrium. Water is a liquid. Remember, liquids and solids are not included within an ICE chart. Here, we're told to find the original molarity, so that's our formal concentration.

We don't know what it is, so we'll just place \( f \) here. These initially are 0. Remember, we're losing our reactant over time in order to produce products. This will be minus \( x \) plus \( x \) plus \( x \). Bring down everything, \( f - x, x \).

Knowing this, we set up the equilibrium expression as \( K_a = \frac{A^- \times H_3O^+}{\text{Reactant}} \). So it's \( K_a = \frac{A^- \times H_3O^+}{\text{Reactant}} \). Again, we're ignoring water because it's a liquid. Our \( K \) is told to be \( 4.7 \times 10^{-3} \). At equilibrium, both of my products are equal to \( x \). So if they're multiplying each other, that's going to be \( x^2 \) on top divided by \( f - x \).

What we have to do here is to figure out that formal concentration. They tell us the pH of the solution. Remember, if we know pH, then we know \( [H_3O^+] \) concentration. With that \( x \) variable, you can plug it into here and plug it into here. We'll have all the variables we need and the only missing one will be our formal concentration which we'll have to solve for.

Alright, so we're going to plug in \( 10^{-4.12} \). When we do that, that gives me an answer of \( 7.59 \times 10^{-5} \). So again, this is equal to our \( x \) variable. So what we're going to have here is \( 4.7 \times 10^{-3} = \frac{(7.59 \times 10^{-5})^2}{f - 7.59 \times 10^{-5}} \).

So all we need to do again is figure out our formal concentration. So cross multiply these two. So this will distribute into here and distribute into here. So when we do that, that's going to give me at this point \( 4.7 \times 10^{-3} \times (f - 3.56531 \times 10^{-7}) = 5.7544 \times 10^{-9} \).

Remember, we need to isolate our formal concentration. So we're going to add \( 3.5631 \times 10^{-7} \) on both sides. So this cancels out. Bring down \( 4.7 \times 10^{-3}f = 3.62286 \times 10^{-7} \). Isolate your formal concentration, so divide both sides by \( 4.7 \times 10^{-3} \).

So here at the end, our formal concentration equals \( 7.71 \times 10^{-5} \) molar. That there represents our original or formal concentration of our weak acid. Remember, things to look out for. If we have a weak acid, we'll have a \( K_a \) less than 1. If we have a weak base, we'll have a \( K_b \) less than 1. Here, if they're telling us pH, that can give us \( H_3O^+ \). If they gave us pOH, then that will give us \( OH^- \). These are key things that you need in order to figure out whether they're asking for the concentration of the original weak acid or one of its products.

Weak Acid-Base Equilibria Calculations 1

Here it says you are seeking to identify an unknown monoprotic acid by determining its Ka value. Here we're told a 6.05×10-2 molar solution of this unknown monoprotic acid has a pH of 2.122. Determine the Ka of this unknown acid. All right. We know that we're dealing with a weak acid in this case because strong acids have Ka values that are equal to infinity because they're so much greater than 1.

Our generic form of a weak acid is weak acids, weak bases react with water in solution. Because it's the acid, we know that it would donate an H⁺ to water and therefore create A^- plus H₃O⁺. We know that we'd be dealing with some type of ice chart to show this equilibrium process. Water is a liquid. Liquids and solids are not included within an ICE chart.

The initial concentration is 6.05×10-2 molar. Initially, we don't have anything for the products, so they're both 0. We lose reactants in order to create products. Bring down everything. 6.05×10-2-X+X+X.

Now we're looking for Ka. Ka is equal to products over reactants. The setup is quite similar to the example we saw above. So A^- times H₃O⁺ divided by HA. Remember, we're looking for Ka now.

So at equilibrium, both our products are X, so that's x2/6.05×10-2-x. Like above, if we know pH, that means we know the concentration of H₃O⁺ because H₃O⁺ is equal to 10-pH. When we do that, that's going to give me 7.5509×10-3. And again, this equals X like we set up above. So plug that number in for X.

So 7.5509×10-32/6.05×10-2-7.5509×10-3. When we plug that in, what you'll get as your answer is 1.08×10-3 for my Ka value. Remember, equilibrium expressions, Ka, Kb, they don't have any units, so it's just that number. Looking at our options, option D would have to be the correct choice. Remember, both questions on this page, example 1 and 2, although asking different things, the setup is quite similar.

When dealing with a weak acid or a weak base, we have an equation that's occurring at equilibrium. With this, we set up an ice chart just to help us organize the equilibrium expression for Ka so that we can know that K equals products over reactants and then just solve for the missing variable. In this case, the missing variable was our Ka value.

Weak Acid-Base Equilibria Calculations

Here it tells us a weak acid has a \( K_a \) of 5.35. What is the hydronium ion concentration in a 0.10 molar solution of this weak acid? We know that we're dealing with a weak acid, and the general generic formula of a weak acid is HA. Now remember, weak acids and weak bases react with water within the solution. So this weak acid will react with water.

It is the acid, so that means water is the base. Remember, according to theory, the acid donates an \( H^+ \). What we're going to create is \( A^- + H_3O^+ \). We're dealing with an ICE chart because we're dealing with a weak species. Anytime we're dealing with a weak acid or weak base, we set up an ICE chart.

Remember, in an ICE chart, we ignore solids and liquids. This water, which is a liquid, is ignored. Our initial concentration is said to be 0.10 molar. We are not given the initial concentration for our products, so they're both 0. Remember we lose reactants in order to make products.

Bring down everything. Since we're dealing with a weak acid, we want to use our acid dissociation constant or \( K_a \). \( K_a \) is equal to products over reactants. Our products multiply together divided by our reactant on the bottom. Again, we ignore water because it's a liquid.

Here, the issue is we don't have \( K_a \), we have \( pK_a \). So, we're going to have to convert \( pK_a \) into \( K_a \). Here, \( K_a = 10^{-pK_a} \). We plug that \( pK_a \) value given to us of 5.35 in. That gives me \( 4.467 \times 10^{-6} \).

That's my \( K_a \) which I'm going to plug over here. At equilibrium, both my products are x. So \( x \times x = x^2 \) divided by what's on the bottom which is \( 0.10 \, \text{minus} \, x \). Here, we have this minus x. Although this is analytical chemistry, we want to be as precise as possible.

If that x variable is insignificant enough, we can ignore it. To determine if we can ignore it, we do the 5% approximation method. Now, the 5% approximation method says that if I take my initial concentration of my weak species and I divide it by its dissociation constant, which is \( K_a \) in this case, if I divide the initial by the \( K_a \) and the ratio gives me a value greater than 500, then I can ignore that minus x. My initial concentration is 0.10 molar. We're going to divide it by the \( K_a \) we just found.

\( 4.467 \times 10^{-6}\). When I do that, it gives me 22,387.2 as a value. That number is definitely greater than 500. According to the 5% approximation method, I can ignore that minus x there. All we do now is we solve for x.

We're going to multiply both sides by 0.10. We're going to have \( 4.467 \times 10^{-7} = x^2 \). We just want x, so we're going to take the square root of both sides here. So when I take the square root of both sides here, I get \( x = 6.68 \times 10^{-4} \, \text{molar} \). When I solve for x here, this x gives me this x, which is equal to \( H_3O^+\).

Remember, when we're setting up an ICE chart for a weak acid or a weak base, anytime we find x, that represents either my \( H_3O^+ \) concentration or my \( OH^- \) concentration. Since I'm dealing with an acid here, it's going to give me the \( H_3O^+ \) concentration, which is my hydronium ion concentration. Based on our answer, option F would be the correct choice. Just remember some of the fundamentals that we went over concerning this question.

If we're dealing with a weak acid or a weak base, in order to find hydronium ion concentration or hydroxide ion concentration, we'll have to set up some type of ice chart in order to solve for x. Once we do, we'll have that number. If we wanted to go a step further, we could have taken the negative log of this concentration, and that would have given me the pH. Here, we're not asked to find the pH. We're just asked to figure out what the concentration is of the hydronium ion. We just stop there, for x being \( 6.68 \times 10^{-4} \) molar.

The pH of an aqueous 0.10 M nitrite ion is 8.17. What is the base dissociation constant of the base?

• A. 4.6 x 10^-16
• B. 2.2 x 10^-11
• C. 1.6 x 10^-6
• D. 1.6 x 10^‑5
• E. 1.2 x 10^-3

Answer: (B)

Here’s what students ask on this topic:

How do you determine the original molarity of a weak acid given its Ka and pH?

To determine the original molarity of a weak acid with a given Ka and pH, follow these steps: First, calculate the hydronium ion concentration using the formula ${10}^{-\mathrm{pH}}$. For example, if the pH is 4.12, the hydronium ion concentration is ${10}^{-4.12}$, which equals 7.59 × 10^-5 M. Next, set up the equilibrium expression $K=\frac{[A-]\left[H_{3}O\right]}{\left[}$ and solve for the formal concentration. This will yield the original molarity of the weak acid.

What is the significance of the Ka value in weak acid-base equilibria?

The Ka value, or acid ionization constant, is crucial in weak acid-base equilibria as it measures the strength of a weak acid. A smaller Ka value (less than 1) indicates a weaker acid, meaning it partially ionizes in solution. The Ka value helps predict the extent of ionization and the concentrations of the acid and its conjugate base at equilibrium. It is used in equilibrium expressions to calculate pH, pOH, and the concentrations of various species in the solution.

How do you use an ICE chart to solve weak acid equilibrium problems?

An ICE chart (Initial, Change, Equilibrium) is a systematic way to solve weak acid equilibrium problems. Start by listing the initial concentrations of the reactants and products. Then, indicate the changes in concentration as the system reaches equilibrium, using variables like x. Finally, express the equilibrium concentrations in terms of these variables. Use the equilibrium expression (Ka) to set up an equation and solve for the unknowns. This method helps organize the data and simplifies the calculation process.

How do you calculate the pH of a weak acid solution?

To calculate the pH of a weak acid solution, follow these steps: First, write the ionization equation of the weak acid. Use the given Ka value and set up the equilibrium expression. Create an ICE chart to determine the equilibrium concentrations of the species involved. Solve for the hydronium ion concentration [H₃O⁺]. Finally, calculate the pH using the formula $-{10}^{-\mathrm{pH}}$. For example, if [H₃O⁺] is 7.59 × 10^-5 M, the pH is -log(7.59 × 10^-5), which equals 4.12.

What is the relationship between pH and pOH in weak acid-base equilibria?

The relationship between pH and pOH in weak acid-base equilibria is given by the equation pH + pOH = 14 at 25°C. This relationship helps in determining the concentration of hydroxide ions (OH^-) if the pH is known, and vice versa. For example, if the pH of a solution is 4.12, the pOH can be calculated as 14 - 4.12 = 9.88. This relationship is essential for understanding the acid-base balance in a solution and for calculating the concentrations of H₃O⁺ and OH^- ions.