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

Weak bases, characterized as weak electrolytes, only partially dissociate in solution, necessitating the use of an ICE (Initial, Change, Equilibrium) chart to calculate equilibrium concentrations in molarity (M). The base ionization constant (K_b) is essential for determining these equilibrium amounts. To assess the extent of ionization, the percent ionization formula is used:

Percent Ionization=[OH-] at equilibrium/Initial concentration of weak base×100

This calculation reveals that weak bases ionize less than 100%, unlike strong bases, which ionize completely. Understanding this concept is crucial for predicting the behavior of weak bases in aqueous solutions and their ionization efficiency.

ICE Charts of Weak Bases

Now recall that weak basis represent weak electrolytes that only partially dissociate into aqueous ions. Here we're going to say that they require the use of an ice chart. So ICE stands for initial change equilibrium and we use this ice chart to calculate equilibrium amounts.

Now here the units of an ice chart will be in molarity, which remember is moles per liter and we use capital M to represent molarity. And since we're dealing with weak bases, we'd use our base association constant which is KB.

So just keep in mind when it comes to weak basis, because they're weak species, we have to utilize an ice chart in order to tell what their exact equilibrium amounts will be in a given process.

pH of Weak Bases Example

Here it says to calculate the hydroxide ion concentration of a 0.55 molar potassium fluoride solution at 25¬∞C. Here we're told the acid association constant of HF hydrofluoric acid is 3.5×10-4. All right, so step one is we're going to set up an ice chart for the weak base that has it reacting with water. Now we know that this is a weak base because this is an ionic compound, so we break it up into its ions. So we have K positive and F minus. Based on our understanding of ionic salts, we know that potassium is a main group. Metal main group metals need to be +3 or higher in terms of charge in order to be acidic. Since it doesn't meet that minimum requirement, it's neutral F minus.

On the other hand, adding H2 it creates HF, which is a weak acid, which means that this F minus is basic. But since it comes from a weak acid, it is stronger. But it's still relatively weak, right? So F minus here is our weak base. Now for ionic bases, which this one is, we ignore the neutral metal cation. Here we ignore the potassium ion. Now use the Bronsted-lowry definition to predict the product's form. Make sure that KB is used in the presence of the weak base. So right now we are only given the Ka of its acid form, but we're going to need to utilize its KB later on to find our final answer.

So remember we're going to set up our equation so F minus here will react with water, which is a liquid. Since it is a base, water is going to act as the acid. Water would therefore give an H plus over to F minus. This would create HF aqueous plus OH minus aqueous as products. Step 2. Use the initial row. So remember for this I stands for Initial change equilibrium. Using the initial row, place the amount given for the weak base. So the weak base we're told it was .55. Molar places 0 for any substance not given an initial amount in an ice chart. We ignore solids and liquids, so the water is ignored. We're not told anything about our products initially, so they're both 0.

Step three, we lose reactants to make products, so using the change row, place A -, X for the reactants and A + X for the products, so minus X since we're losing it, and plus X for the product since we're making them. We bring down everything, so .55 -, X + X + X. Now, using the equilibrium row, set up the equilibrium constant expression with AB again 'cause it's a weak base and solve for X. We checked if a shocker can be utilized to avoid the quadratic formula. Alright, so the shortcut that we're going to utilize is called the 500 approximation method. Basically, we're going to take the ratio of initial concentration of our weak base and divide it by our KB value. If we get a number greater than 500, then we can ignore the minus X within our equilibrium expression, right?

So here we're going to say kW equals KA times KB. We need to isolate KB, so KB equals kW divided by ka. So this would be 1.0×10-14 divided by our ka, which is 3.5×10-4. Doing that gives us a KB value of 2.857×10-11. Alright, so our initial concentration is 0.55 molar. We found our KB as 2.857×10-11. When we punch that in, we get 1.925×1010. So we get a number much greater than 500, which means we'll be able to ignore the minus X within our equilibrium expression.

Remember your equilibrium expression for this weak basis KB equals products over reactants, so we plug in the products over reactant. Water is a liquid so we ignore it. So bringing down this expression and inserting values for what we have AB we said is 2.857×10-11. Both the products are X at equilibrium, so multiply together as X ^2 and this will be oh .55 -, X. Well remember we just did the ratio of initial concentration divided by KB and we saw a number much greater than 500, which means we can ignore this minus X and avoid the quadratic formula that we have here.

So then all it becomes is cross multiplying my KB with .55. So I'm going to do that up here. So when I do that, I'm going to hit X ^2 = 1.57135×10-11. I don't want X ^2, I just want X. We'll take the square root of both sides X = 3.96×10-6 molar. Now this is our answer because we're asked to find the hydroxide ion concentration, and at equilibrium that's equal to X. So we just found out what my X value is, which is equivalent to my hydroxide ion concentration. So my final answer will be 3.96×10-6 molar.

pH of Weak Bases Example

Here we can say that the H Oregon OH of a weak base can be calculated once the equilibrium concentration of the hydroxide ion is found. Now this is determined by using the equilibrium row of an ice chart. In this example question, it says what is the pH of a 0.12 molar ethyl amine solution. We're told that the K_B value of ethylamine is 5.6×10-4.

All right, so we're going to start out by following the steps one to three and setting up an ice chart. So here we have our ethylamine. The ethylamine, being a weak base because its K_B is less than one, would react with liquid water. Since it's the base, water will act as the acid, donating an H to it. This H⁺ would go towards the nitrogen, giving us C₂H₅NH₃⁺. This is the ethyl ammonium ion. Losing an H would become OH^- ion.

We have our equation and this is a nice chart which is initial change equilibrium. For our initial row, we're going to place the initial amount given for ethylamine, which is 0.12. Molar liquids and solids are ignored within a nice charge, so water will be ignored. Our products initially are zero since they haven't formed yet, and we're losing reactants minus X. To make products plus X + X we bring down everything, so we have 0.12 - X + X + X.

This leads us to Step 4 where using the equilibrium row, set up the equilibrium constant expression and solve for X. Here our equilibrium constant expression. Since this is a base, we have K_B equals products over reactants, so it equals ethyl ammonium ion times hydroxide ion over ethyl amine. Here we check to see if a shortcut can be utilized to avoid the quadratic formula. The shortcut that we utilize is called the 500 approximation method.

In it, we take the ratio of our initial concentration of our weak base, in this case to our K_B value, and if that ratio happens to be greater than 500, we can ignore the minus X in terms of our equilibrium constant expression. So here we have our initial concentration of 0.12 molar divided by the K_B value of the weak base. When we do that, it only gives me 214.3 as a value. Since the number is not greater than 500, we cannot ignore the minus X, which means we'll have to utilize the quadratic formula, which again remember is -b±b2-4ac/2a.

So come up here. Our expression is 5.6×10-4 for the weak base and that be x2 / 0.12 - X. Cross multiply those so when we do that we're going to get 5.6×10-4 and in ( 12 - X and that equals x2. So that give me 5.6 times. Oops, 5.6×10-4 * 12 gives me 6.72×10-5 - 5.6×10-4 X, and this equals x2.

Now my 2X variables, the x2 1 have the larger power, so it's our lead term. So you would add 5.6×10-4 X to both sides. You need subtract 6.72×10-5 on both sides. When we do that, we have our new equation, so x2 + 5.6×10-4 X - 6.72×10-5. Here this equation would equal AB and C Setting U our quadratic formula, we'd have -5.6×10-4±5.6×10-4,4×1×-6.72×10-5/2×1.

When we solve for this, realize that we're going to figure everything in here, and then take the square root of that answer. When we do that, we get -5.6×10-4±0.0164/2. So here there are two outcomes for X because it's plus or minus. So X here could equal -5.6×10-4+0.0164/2 or X could equal -5.6×10-4-0.0164/2. This gives us 2 answers for X initially. So in the first top when we get 0.00792 molar and in this bottom one we get negative 0.00848 molar.

But which one is the correct one? The correct X is the X that wherever you place it in terms of the equilibrium row, you're going to get a positive answer. That means it's -1 will not work because it's -1. If I placed it here or here would give me a negative concentration at equilibrium, which is not possible. Alright, so here the X that we've just found takes us to Step 5. The X variable that we just found would give us our hydroxide ion concentration.

Since we know hydroxide ion concentration, we could find pOH. So we take that number that we just found. We plug it in here 0.00792. That would give me 2.10 as my pOH and then remember that pH equals negative log of H⁺ and that would give you my pH. Or you could also remember that pH plus pOH equals 14, so pH equals 14 minus pOH. O here would be 14 - 2.10, and if it's 2.10 that come out to 11.90, so we'd have 11.90 as our pH for this solution.

OK, so just remember when we find X, look on the equilibrium row that X will give you either H₃O⁺ or in this case OH^-. You just got to check simply give this OH^- we find OH and from there subtract it from 14 to find H.

Calculating Percent Ionization

When it comes to calculating the percent ionization or percent association, it's important to remember that weak bases also represent weak electrolytes that only partially ionized or dissociate into aqueous ions. Because of this fact, weak basis ionize less than 100%.

Strong bases, on the other hand, they represent strong electrolytes and therefore ionize completely. So they would ionize 100%. If we wanted to determine the percent ionization or percent association for a weak base, we utilize the following formula.

Here we're going to say percent ionization and percent dissociation equals the concentration of hydroxide ion at equilibrium divided by the initial concentration of my weak base times 100. Utilizing this formula will be able to determine the percent ionization of any given weak base.

Percent   ionization   =   Concentration   of   hydroxide   ion   at   equilibrium Initial   concentration   of   weak   base   ×   100

pH of Weak Bases Example

Calculate the percent ionization when 73.2g of sodium hypoiodite are dissolved with 500 MLS of solution. We're told that the K value of hypoiotis acid, which is Hio, is 2.3 * 10^-11. All right, so we're going to say that hypoiotis acid is a weak acid. Since its KA is less than one. Naio represents its conjugate base. It has one less hydrogen because it's the conjugate base. It represents the weak base O.

We're going to have to use the steps from 1:00 to 3:00 to set up our ice chart for this weak base. Remember, for ionic bases we ignore the neutral metal cation, in this case the sodium ion. This leaves us with the hypoiodite ion alone. Since it's a base, it will react with water. Water here will be in its liquid phase. It is the base, so water acts as the acid, meaning that water will donate an H plus to it. This creates Hio plus OH minus as products. We're dealing with a weak base, so this is our ice chart setup.

So we have initial change equilibrium. Now remember, the units of an ice chart with a weak base have to be in molarity. So we're going to take the 73.2g of sodium iodide and we're going to convert that into molds. So we're going to stay here. For every one mole of sodium iodide, we have a mass of 165.89 grams of sodium hypoiodite. That's the mass of the sodium, the iodine and oxygen together. So here grants cancel out and we're going to have here for everyone mole. We have a sodium iodide. We have one mole of iodide ion which equals .441 moles.

We need molarity for the ice chart, so molarity here will be moles over liters should be 0.441 moles divided by .500 liters, which comes out to .882 molar. Now with the ice chart, we ignore solids and liquids, so the water would be ignored. We don't know anything about the products initially, so there's zero. We lose reactants in order to make product, bring down everything. Now here, using the equilibrium row, set up the equilibrium constant expression and solve for X. Here we'll check if a shortcut can be utilized, and if so, we can avoid the quadratic formula.

Since this is a weak base, we're going to have to utilize KB. KB would equal products over reactants. However, we're not given KB initial, we're giving Ka. So we have to remember that KA times KB equals kW. So we're gonna say KB here equals kW divided by ka. So here this would be 1.0 * 10^-14 divided by RKA, which we're told is 2.3 * 10^-11. Doing this gives our KB value. Our KB value here would equal 4.3 * 10^-4.

Now that we know the KB value, we can do the 500 approximation method when the ratio of the initial concentration divided by, in this case, KB. If the ratio happens to be greater than 500, then we'll be able to ignore the minus X within our equilibrium expression. So our initial concentration is .882 molar, and again our KB is 4.3 * 10^-4. When we punch that in, that gives us 2051.2, so we got a number greater than 500. So we can ignore the minus X within our equilibrium expression.

So here's our equilibrium expression when we plug in the values and. Again, because the ratio is greater than 500, I can ignore this minus X. So coming up here we can say this is X² divided by 0.882. By ignoring the minus X we can avoid the quadratic formula. We're going to cross multiply these two, so when we do that we're going to X² = 3.7 times 10^-4 and then take the square root of both sides. X will give me approximately 0.01947 when we find X.

Here X gives me OH concentration, so that takes us to step five. We would use the X variable which we just found to help us calculate the percent ionization or dissociation. So remember here we're going to say percent ionization or percent association for a week base equals the the concentration of hydroxide ion at equilibrium, which we find is .01947 molar UMM divided by the initial concentration of my base which is 0.882 molar times 100. So when we do that we get 2.21%. So this will represent the percent ionization of my hypoiodite ion.

Determine the pH of a solution made by dissolving 6.1 g of sodium cyanide, NaCN, in enough water to make a 500.0 mL of solution. (MW of NaCN = 49.01 g/mol). The K_a value of HCN is 4.9 × 10⁻¹⁰.

• A. 2.648
• B. 13.389
• C. 5.294
• D. 11.352

An unknown weak base has an initial concentration of 0.750 M with a pH of 8.03. Calculate its equilibrium base constant. 

• A. 9.35 × 10⁻⁹
• B. 1.53 × 10⁻¹²
• C. 6.54 × 10⁻³
• D. 1.07 × 10⁻⁶

Here’s what students ask on this topic:

How to find the pH of a weak base?

To find the pH of a weak base, you'll need to follow these steps:

1. Identify the weak base and its concentration in the solution. Let's say the base is B and its concentration is _[B].
2. Write down the equilibrium expression for the base dissociation: B + H₂O ⇌ BH⁺ + OH^-.
3. Use the base dissociation constant (K_b) for the weak base, which can be found in tables or literature. The K_b expression is K_b = ^[BH+][OH-]/_[B].
4. Set up an ICE table (Initial, Change, Equilibrium) to calculate the changes in concentrations of the reactants and products at equilibrium.
5. Assuming x is the change in concentration of OH^- ions at equilibrium, the K_b expression becomes K_b = ^x2/_{[B] - x}. Since weak bases only partially dissociate, x is much smaller than [B], allowing the approximation [B] - x ≈ [B].
6. Solve for x, which represents the [OH^-] at equilibrium.
7. Calculate the pOH and then use the relationship pH + pOH = 14 to find the pH.

What pH is a weak base?

A weak base is a base that does not fully dissociate into its ions in an aqueous solution. As a result, weak bases have a pH that is greater than 7 but less than 14. The exact pH of a weak base solution depends on the base's concentration and its dissociation constant (K_b). Typically, weak bases have a pH ranging from about 7.1 to just below 11. Examples of weak bases include ammonia (NH₃), with a pH around 11 when in a standard 0.1 M solution, and methylamine (CH₃NH₂). It's important to note that the pH can vary significantly with concentration and temperature. Weak bases partially accept protons, resulting in an equilibrium between the base and its conjugate acid in solution, which is why they do not achieve the higher pH levels of strong bases, which fully dissociate.

What is a weak base on the pH scale?

A weak base is a substance that does not fully ionize in solution, meaning it does not completely dissociate into its ions in water. On the pH scale, which ranges from 0 to 14, a weak base typically results in a pH value above 7 but below 14. The exact pH value will depend on the concentration of the base and its dissociation constant (_Kb). Weak bases partially accept protons from water, leading to the formation of the hydroxide ion (^OH-) and the corresponding conjugate acid of the base. Because they do not fully ionize, the solution will have a lower concentration of hydroxide ions compared to a strong base at the same concentration, resulting in a less alkaline pH. Examples of weak bases include ammonia (_NH3), methylamine (_CH3NH₂), and ethylamine (_C2H₅NH₂). The pH of a weak base solution is often found using the Henderson-Hasselbalch equation or by calculating the hydroxide ion concentration and then using the pOH to find the pH.