pH of Weak Acids - Online Tutor, Practice Problems & Exam Prep

Weak acids are classified as weak electrolytes, partially ionizing in aqueous solutions, unlike strong acids which ionize completely. To determine the equilibrium concentrations of a weak acid, an ICE (Initial, Change, Equilibrium) chart is used, with molarity as the unit of measurement. The acid dissociation constant, Ka, is essential for these calculations. The percent ionization of a weak acid, which is less than 100%, indicates the extent of its ionization and is calculated using the formula:

$\left(\frac{\text{equilibrium concentration of}{H}_{3}O+}{\text{initial concentration of the weak acid}}\right)\times 100$

This calculation requires knowing or determining the hydronium ion concentration at equilibrium, often through an ICE chart. Understanding these concepts is crucial for analyzing acid behavior in solutions.

ICE Charts of Weak Acids

Now recall that weak acids represent weak electrolytes that only partially dissociate into ions. Here we're going to say they require the use of an ICE chart. Now, ICE here stands for Initial change equilibrium, so they require the use of an ICE chart in order to calculate equilibrium amounts.

We're also going to say that the units of an ICE chart for a weak acid will be in molarity, and because it's a weak acid we'll use K_a which is our acid association constant. So keep this in mind more as to find the equilibrium amounts for any potential weak acid.

pH of Weak Acids Example

Here in this example question, it says calculate the hydronium ion concentration for 0.30 molar of hydrocyanic acid. The acid association constant K_A for hydrocyanic acid is 4.9×10-10. All right, so what we need to realize first is that hydrocyanic acid is a weak acid and we know this because its K_A is less than one. Remember, if your acid has K_A less than one, it is a weak acid and they're asking us to calculate the hydronium ion concentration, which is H⁺ concentration or H₃O⁺ concentration.

To do that, to calculate this equilibrium amount, we follow the following setup. So step zero. We use the following steps when asked to determine the equilibrium concentration of any compound in your equation. Remember the brackets there mean concentration here. Step one, we set up an ice chart for the weak acid that has it reacting with water. So here goes our hydrocyanic acid. We're going to react it with water here. Water will be in its liquid phase. Now use Bronsted-Lowry definition to predict the products formed. Make sure that K_A is used in the presence of weak acid. We don't have to worry about that because they gave us K_A in the very beginning.

Using the Bronsted-Lowry definition, remember that your hydrogen cyanic acid is an acid and that means your water will act as a base. Remember the acid donates an H⁺, so when it does that, we're going to get CN^- being created. Since it's an ion, it's aqueous. Water accepts an H to become H₃O⁺ also aqueous. So here goes our equation for the dissociation of our weak acid. Now here we've set up an ice chart. So we're going to say we have initial change equilibrium. Now using the initial role, place the amount given for the weak acid. Place as zero for any substance not given an initial amount.

All right, so we're told within the question that we're dealing with 0.30 molar of hydrocyanic acid. Remember, in an ice chart, we ignore solids and liquids. Water is a liquid, so water will be ignored. We're not told anything about the initial amounts of our products, so they're both initially 0. Step three, we lose reactants to make products. So using the change row place A -, X for the reactants and 3rd we're losing them and A + X for the products since we're making them. So this will be minus X + X + X. Now using the equilibrium row, set up the equilibrium constant expression with K_A. So remember we're using K_A and solve for X.

Now here for the equilibrium row we just bring down everything. So this is going to be 0.30 -, X plus X + X. Now here we're going to check to see if a shortcut can be utilized to avoid the quadratic formula here. To do that, we're going to use what's called the 500 approximation method. In it, we're going to use the ratio of initial concentration to the K that we're using in this question, in this case would be K_A. And if that ratio is greater than 500, then we could ignore the minus X within our equilibrium expression. So here our initial concentration given to us for hydrocyanic acid within the question was 0.30 molar. The K_A of this weak acid is 4.9×10-10.

When we punch that in, it gives me a number of 6.12×108. So this number is way larger than 500, which means I can ignore the minus X within my equilibrium expression. Remember your equilibrium expression? We come back up here. Remember that is K_A equals products over reactants, so it equals CN^- times H₃O⁺ divided by HCN. Again, water is a liquid. We ignore solids and liquids, that's why it's not there if we plug in the numbers that we want for it. So K_A is 4.9×10-10. Both of my products are X, so multiply together that would be X². And then at equilibrium my HCN is 0.30 -, X.

By using the 500 approximation method we saw that our ratio was greater than 500, so I could ignore this minus X and because of that I can ignore the quadratic formula. But just for reference, remember the quadratic formula here would be -b ± b² -, 4 AC over 2A. So this would be our quadratic formula. Luckily we don't have to use it, so plug in what we know. We have 4.9×10-10 = X² / 0.30. We're going to cross multiply these two together. When we do that, that's going to give me X² = 1.47×10-10. We want X, not X². So take the square root of both sides here. So when I do that, that's going to give me X, which is going to give me 1.2×10-5 molar.

Now what is the question asking you to find? It's asking me to find the concentration of hydronium ions. When we say that we're really talking about hydronium ions at equilibrium. Look at the ice chart here. Is my hydronium ion. At equilibrium it's equal to X, which is what I just found. So this number represents the hydronium ion concentration at equilibrium. So 1.2×10-5 molar would be my answer.

pH of Weak Acids Example

Now the pH or POH of a weak acid can be calculated once the equilibrium concentration for the hydronium ion is found. Here we're going to say that this is determined by using the equilibrium row of the ice chart here. In this example question it says what is the pH of a 0.074 molar nitrous acid HNO₂ solution. The K value for the compound is 4.6×10-4. Since it's K_a is less than one, we know that nitrous acid is a weak acid and because of that we're going to have to utilize an ice chart.

Now here we're going to use steps one to three to set up the ice chart. So first of all, we're going to have our weak acid react with water. Warrior will be in its liquid phase. Since nitrous acid is an acid, that means that water will act as a base. Remember, based on the Brønsted-Lowry definition, acids donate H⁺ bases accept H⁺. Doing this would create our nitrite ion plus the hydronium ion. Here we have our balanced chemical equation and we're going to say since we're using an ice chart that stands for initial change equilibrium.

When it comes to this ice chart, we're going to plug in the initial amount given for our weak acid, which is 0.074 molar. Remember that liquid and solids are ignored within an ice chart, so these would get X. We're not told anything about our products initially, so they're zero for the changegrown. Remember we lose reactants to make products, so my react would be minus X, products will be plus X. Step 4, using the equilibrium row, set up the equilibrium expression and solve for X. For the equilibrium row we bring down everything, so this would be 0.074 - X + X + X here.

At this point, when we use the equilibrium constant expression, we have to check with our shortcut to see if we can avoid the quadratic formula. Doing this, we use the 500 approximation method. We're going to say here when the ratio of the initial concentration divided by R K_a. In this case, if it's a ratio greater than 500, we can ignore the minus X within our equilibrium expression. Here our initial concentration of the weak acid is 0.074 molar. Its K_a is told to us as 4.6×10-4. When we put this into our calculators we get back 160.87. So unfortunately we don't have a number greater than 500 which means we cannot ignore the minus X and therefore we have to set up a quadratic formula.

Now here in this section we have our equilibrium expression with the values plugged in. But coming back up here, remember that your equilibrium expression is K_a equals products over reactants. Again we ignore solids and liquids. K_a again is 4.6×10-4. At equilibrium, both our products are X, so multiply together their X² divided by 0.074 - X. Ross, multiply these together. So we're going to come over here. So it's going to be 4.6×10-4 and in (0.074 - X, which equals X². We're going to distribute, distribute. So when I do that, I'm going to get here 3.404×10-5 - 4.6×10-4 X = X² here are both of our X variables, but the X² is the one with the larger power, so it's our lead term, which means we bring everything over to the left side, actually to the right side to the right side.

So we're going to add 4.6×10-4 X to both sides, and we're going to subtract 3.404×10-5 from both sides. So when we do this, we're gonna get our equation. Our equation now is going to be X² + 4.6×10-4 X - 3.404×10-5. This would be our A, our B and our C. This is important so that we can plug it into the quadratic formula. So now we have -4.6×10-4 plus or minus the square root of 4.6×10-4² - 4. A here would just be one C. Don't forget the negative sign divided by 2 * 1.

Now here when we do the square root of everything within here, we're going to get X = -4.6×10-4 ± 0.011678 / 2. Now here it's important to realize that this is plus or minus, which means we're going to get 2 answers where X = -4.6×10-4 plus 0.011678 divided by two, and one where it's 4.6×10-4 minus 0.011678 / 2. OK, so there's two possibilities. You're either adding or subtracting this number. When we see that we're going to get 2X values. 1X will come out to be 0.005609 molar. The other one will come out to be -0.006069 molar.

Only one of them is the right answer. How do we know which one is the right answer? Well, you would take your X value and no matter where you plug it in terms of the equilibrium row, you should get a positive answer as a result. Because of this, we know that this second one that will not work because if I place that negative X here or here, I'd have a negative concentration at equilibrium, which is not possible. OK, you cannot have a negative value at equilibrium. So that means the X variable is this value here. And if we look, this X variable that we just found gives us H₃O⁺. That leaves us to Step 5.

So for Step 5, the X variable that we just found gives us the hydronium ion concentration and it can be used to solve 4 pH. Remember pH equals negative log of the hydronium ion concentration, which again we said is 0.005609. And then here when we plug that in we get 2.25. 2.25 would be the pH or nitrous acid solution.

Calculating Percent Ionization

Now, percent ionization or percent association represents the percentage of a weak acid that can become ionized when placed in an aqueous solution. Remember that weak acids represent weak electrolytes and strong acids represent strong electrolytes. Because of this, weak acids ionize less than 100% when placed in an aqueous solution and strong acids ionize 100% when placed in an aqueous solution.

When we want to calculate the percentage of ionization or the association for a weak acid, we utilize the following formula: the percent ionization or percent dissociation of a weak acid equals the equilibrium concentration of H3+ divided by the initial concentration of the weak acid times 100.

Now, in order to determine the concentration of hydronium ion at equilibrium, they either give it to you or you need to calculate it using an ICE chart. So keep this in mind when you need to figure out the percent ionization or percent dissociation of any given weak acid.

pH of Weak Acids Example

Calculate the percent association of 4.10 * 10 to the -1 molar of acetic acid. Here we're told that the KA value for acetic acid is 1.8 * 10 to the -5. All right, since it's a weak acid, we need to set up an ice chart, and we know it's a weak acid because its KA is less than one. So here we're going to use steps one to three to set up the ice chart.

So first, we're going to react the weak acid with water. In this case, water will be in its liquid form following the Bronsted-Lowry acid definition that we know. We know that the acid is acetic acid and therefore the water would be the base. The acid donates an H plus and the basic subset. Doing this would create as our products the acetate ion and the hydronium ion.

This is an ice chart. So which stands for initial change equilibrium. Now we place the initial amount given to us for the weak acid, which we're told is 4.10 * 10 to the -1, which is just basically 0.410 molar. Remember, ice charts ignore solids and liquids, so the water will be ignored. We're not told anything initially about our product, so initially there's zero for the change row. Remember, we lose reactants in order to make products, so minus X + X + X. For the equilibrium row we bring down everything, so .410 -, X + X + X.

So we filled out our ice chart. Now using the equilibrium row, set up the equilibrium constant expression and solve for X. Check if a shortcut can be utilized to avoid the quadratic formula. So here we're going to say what the quadratic formula we have it. Here we say we use the 500 approximation method. When the ratio of initial concentration to the KA value in this case is greater than 500, we can ignore the minus X.

So our initial concentration of acetic acid is 4.1 * 10 to -1 molar. Its KA value was 1.8 * 10 to the -5. When we punch that in, it gives us 22,777.8, a number much greater than 500. That means that our in our equilibrium expression we can ignore the minus X. So remember our equilibrium expression is products over reactants, so it's ka equals my two products divided by the weak asset. Again, water is a liquid so we ignore it.

So 1.8 * 10 to the -5 = X ^2 divided by. Remember using the 500 approximation method below, we saw that we can ignore the minus X. So this is just .410 on the bottom and we didn't include the minus X here. Cross multiply these two When we do that X ^2 = 7.38 * 10 to the -6. Take the square root of both sides X = 2.717 * 10 to the -3. We just found out what X is and it says use the X variable.

Now to calculate the percent ionization association. Here we're going to say that the X that we found is equal to my hydronium ion concentration. So we just found out what H3O plus concentration is. Remember for weak acid, percent ionization or dissociation equals the concentration of H3O plus at equilibrium divided by the initial concentration of the acid times 100. So at equilibrium HDL plus was 2.717 * 10 to the -3 molar. Our initial concentration was 0.410 molar and then multiply that by 100. When we do that, we get oh .66%. So that's the amount of acetic acid that basically ionized when placed in an aqueous solution. It's a weak acid, so we should expect a number that's much smaller than 100%. In this case it .66%.

Calculate the [H⁺] of a 0.50 M solution of methylammonium bromide, CH₃NH₃Br. The K_b of methylamine, CH₃NH₂, is given as 4.4 x 10^-4.

• A. 0.50 M
• B. 1.14 × 10⁻¹¹ M
• C. 3.4 × 10⁻⁶ M
• D. 2.97 × 10⁻⁹ M

An unknown weak acid has an initial concentration of 0.55 M. What is the pH of the solution if the weak acid also has a pK_a of 5.79?

• A. 0.60
• B. 6.05
• C. 3.02
• D. 5.75

Here’s what students ask on this topic:

What are the assumptions when calculating the pH of weak acids?

When calculating the pH of weak acids, several assumptions are made to simplify the process:

1. The acid is only partially ionized in solution, meaning it does not dissociate completely as strong acids do.
2. The concentration of the acid at equilibrium is approximately equal to the initial concentration since the degree of ionization is low.
3. The change in concentration of the acid during ionization is so small that it can be ignored. This is because the amount of acid that dissociates is much less than the amount that remains undissociated.
4. The concentration of hydronium ions ($H_{3}O+$) produced by the ionization of the acid is equal to the concentration of the conjugate base formed.
5. The contribution of hydronium ions from the autoionization of water (2H2O⇌H3O++OH−) is negligible compared to the concentration of hydronium ions from the weak acid.

These assumptions allow the use of the simplified version of the acid dissociation constant (Ka) expression to solve for pH, which is practical for weak acids with a Ka much less than 1.

How do you calculate the pH of weak acids?

To calculate the pH of a weak acid, you'll need to use the acid dissociation constant (Ka) for the acid in question. Here's a step-by-step guide:

1. Write down the equilibrium expression for the dissociation of the weak acid (HA):
   $\mathrm{HA}\rightleftharpoons H^{+}+A^{-}$
2. Write the expression for the acid dissociation constant (Ka):
   $Ka=\frac{{\mathrm{[H]}}^{+}{\mathrm{[A]}}^{-}}{\mathrm{[HA]}}$
3. Set up an ICE table (Initial, Change, Equilibrium) to determine the concentrations of the species at equilibrium. Assume the initial concentration of H+ and A- is zero and the concentration of HA is the initial concentration of the acid.
4. Since the acid is weak, assume that the change in concentration of HA is small and can be approximated as x, where x is the concentration of H+ and A- at equilibrium.
5. Plug the equilibrium concentrations back into the Ka expression and solve for x, which is [H+].
6. Calculate the pH by taking the negative logarithm (base 10) of the H+ concentration:
   $\mathrm{pH}=-\log \left[H^{+}\right]$

Remember that if the percent ionization is greater than 5%, the assumption that the change in concentration of HA is small may not be valid, and a more accurate method may be needed.