Main content

### Course: AP®︎/College Chemistry > Unit 8

Lesson 2: pH and pOH of strong acids and bases# Strong base solutions

Strong bases (such as Group 1 and 2 metal hydroxides) dissociate completely in water to produce hydroxide ions. The concentration of OH⁻ in a strong base solution can therefore be determined from the initial concentration of the base and the stoichiometry of the dissolution. For example, the strong base Ba(OH)₂ dissociates to give two OH⁻ ions per formula unit, so a 0.1

*M*Ba(OH)₂ solution has an OH⁻ concentration of 0.2*M*. Created by Jay.## Want to join the conversation?

- I don't quite understand at4:51why all of the hydroxide ions come from the strong base CaOH2. I mean, doesn't the autoionization of water also play a role in creating OH-? It seems to me that we are underestimating the concentration of OH-(2 votes)
- Technically the hydroxide concentration is due to both the calcium hydroxide and the autoionization of water. However, they have vastly different contributions. The calcium hydroxide delivers a hydroxide concentration of 0.0020 M, while water’s autoionization provides only 1.0 x 10^(-7) or 0.00000010 M. So 0.0020 M + 1.0x10^(-7) M = 0.0020007 M. And when we take the pOH it becomes 2.698818027861, which rounded for sig figs still becomes 2.70 like the pOH in the video. So since water’s autoionization contribution is so small that it’s not even noticeable in the final answer, it’s simply omitted since its inclusion is insignificant compared to calcium hydroxide’s contribution.

Hope that helps.(8 votes)

- Can we always assume the temperature is 25 C unless told otherwise?(3 votes)
- Yep, pretty much.(4 votes)

- At2:50, how come the final concentration for OH- is to two significant figures, when all inputs were to three significant figures?(3 votes)
- Different mathematical operations have their own rules for sig figs. Here Jay is using logs and antilogs which have different rules compared to multiplication and division. Antilogs, which the last operation Jay uses before he gets the hydroxide concentration, states that an answer should have as many sig figs as the mantissa (the decimal portion) of the number being antiloged. So if we're taking 10^(-1.00) and the mantissa are the two 0s then the answer should only have two sig figs then.

Hope that helps.(3 votes)

- How can we use mental math to approximate the value of -log (0.0020)? I understand how 2 in 2.70 was deduced since 0.002=2 x 10^-2 but I don't understand how to calculate the 0.7 without a calculator (i.e: how to guesstimate log 2)?(2 votes)
- In example 2, if we know the concentration of Ca2+, why do we not use the formula pH = -log[H+]? I know it gets the wrong answer but I'm wondering why this isn't a possible method.(1 vote)
- I'm still learning this also but I'm pretty sure it's because you have to use hydronium ions to calculate pH, so a different species of ion wouldn't work. In this case, there's no hydronium ions produced so you can't use anything to calculate pH other than finding the pOH like he did in the video.(2 votes)

- i don't get how we got 2.7 for the POH(1 vote)
- What do you mean? So you have Ca(OH)2 which dissociates into two OH moles and one Ca mole. Since you have the rxn moles, you can set up a mole ratio. We are given that the concentration of Ca(OH)2 is 0.001M. Based on the stoichiometry of the equation, we know that the ratio of Ca(OH)2 and OH is 1:2. So if we have 0.001M Ca(OH)2, we have 2 times that for OH, or 0.002M. Next: the pOH is equal to -log[OH]. We just found [OH] to be 0.002M. So pOH=-log(0.002), or the pOH=2.7. From here, we can use the equation pH+pOH=14 to get that the pH is 11.3. I'm a student as well so apologies if I get anything wrong, but that's pretty much it. Hope this helps(2 votes)

- Why is KOH a solid whereas in the simplified equation of the strong acid in the previous video, HCL was aqueous. Are states of matter significant for Acid/Base reactions? Do they change based off if the rxn is Acid/Base?(1 vote)
- Potassium hydroxide, KOH, is a solid at ambient conditions. We represent this with the (s) symbol (i.e. KOH(s)). When in an aqueous solution, a solution where water is the solvent, potassium hydroxide dissociates into potassium and hydroxide ions. We represent this with the (aq) symbol (i.e. K^(+)(aq) and OH^(-)(aq)).

When HCl is pure and in ambient conditions, it is a gas and is referred to as hydrogen chloride with the symbol HCl(g). When dissolved in an aqueous solution it is referred to as hydrochloric acid with the symbol HCl(aq). Hydrochloric acid is a strong acid so it almost completely dissociates and is sometimes written in its dissociated form, H^(+)(aq) + Cl^(-)(aq).

Most acid-base chemistry you’ll encounter will be with dissociated species in an aqueous solution, where they have (aq) physical states.

Hope that helps.(2 votes)

- I don't quite understand how he got pH + pOH = 14.00M at01:16.(1 vote)
- Why do they find the pOH using the concentration of the OH and then subtracting it from 14 instead of just using the Ca to find pH?

thank you to anyone who answers this(1 vote)- Calcium doesn’t affect the pH of the solution, only the hydroxide does. This is because hydroxide reacts with hydrogen ions, or more accurately hydronium, and neutralizes into water. As a result, hydroxide decreases the concentration of the hydronium and raises the pH. We can relate the hydronium concentration and pH through the autoionization of water.

The autoionization of water is: 2H2O(l) ⇌ H3O^(+)(aq) + OH^(-)(aq)

Which means we can write the equilibrium expression as: Kw = [H3O^(+)][OH^(-)]

And if we apply –log to both sides the equilibrium expression becomes: pKw = pH + pOH. Where pKw is 14 at 25°C.

So if we don’t know the hydronium concentration directly, we can still calculate it and the pH through the hydroxide concentration and pOH. Knowing the concentration of calcium doesn’t help us find the hydronium concentration or pOH, at best it allows us to indirectly know the hydroxide concentration.

Hope that helps.(1 vote)

- For that last question, would the answer be 11 instead of 11.30 because 0.0010 M has 2 significant figures and not 4?(1 vote)

## Video transcript

- [Instructor] When dissolved in water, a strong base like potassium hydroxide will dissociate completely in solution to form hydroxide ions. Potassium hydroxide is an example of a group 1A metal hydroxide. Other examples include lithium hydroxide and sodium hydroxide. Group 2A metal hydroxides
are also considered to be strong bases. For example, calcium hydroxide is a group 2A metal hydroxide, and so is strontium hydroxide. Let's do a problem with a
group 1A metal hydroxide, sodium hydroxide. Let's say the pH of the solution is 13.00 and our goal is to calculate
the initial concentration of sodium hydroxide. First, we brought out
the disillusion equation. Solid sodium hydroxide
dissociates completely in water to form sodium cations and
hydroxide anions in solution. Looking at the balanced equation, there's a one in front of sodium hydroxide and a one in front of hydroxide ions, therefore the concentration
of hydroxide ions is equal to the initial
concentration of sodium hydroxide. And we can find the
concentration of hydroxide ions in solution from the pH. At 25 degrees Celsius, the pH plus the pOH is equal to 14.00. So we can plug the pH into our equation, which gives us 13.00, plus the pOH is equal to 14.00. So the pOH of the
solution is equal to 1.00, and the pOH is equal to the negative log of the concentration of hydroxide ions. So we can plug the pOH into this equation, which gives us 1.00 is
equal to the negative log of the concentration of hydroxide ions. To solve for the concentration of hydroxide ions in solution, first we move the negative
sign to the left side, which gives us negative
1.00 is equal to the log of the concentration of hydroxide ions. To get rid of the log,
we take 10 to both sides. So the concentration of
hydroxide ions in this solution is equal to 10 to the negative 1.00, which is equal to .10 molar. Because sodium hydroxide is a strong base that dissociates completely in solution to form hydroxide ions, if the concentration of hydroxide ions in solution is .10 molar, so is the initial concentration
of sodium hydroxide. Now let's do a problem with
a group 2A metal hydroxide. Let's say the initial
concentration of a solution of calcium hydroxide is .0010 molar, and our goal is to find
the pH of the solution at 25 degrees Celsius. Calcium hydroxide is a strong base that dissociates completely in solution to form calcium two plus
ions and hydroxide anions, and looking at the mole ratios
in this disillusion equation, there's a one in front
of calcium hydroxide, a one in front of calcium two plus, and a two in front of hydroxide ions. Since the mole ratio of
calcium hydroxide to calcium is one to one, if the initial concentration
of calcium hydroxide is .0010 molar, that's also the concentration
of calcium ions. So it's .0010 molar in solution. The mole ratio of calcium
hydroxide to hydroxide ions is one to two, so if the initial concentration
of calcium hydroxide is .0010 molar, the concentration of
hydroxide ions in solution is twice that concentration, so two times .0010 molar
is equal to .0020 molar. Now that we know the
concentration of hydroxide ions, we can calculate the pH of the solution. One way to calculate the
pH is to first find the pOH of the solution, and pOH is equal to the negative log of the concentration of hydroxide ions. So we can plug in the
concentration of hydroxide ions into our equation, which gives us the pOH of the solution is equal to the negative log of .0020, and when you do the calculation, the pOH is equal to 2.70. Notice, since we have
two significant figures for the concentration, we need two decimal places for our pOH. So to find the pH, we know that pH plus pOH is equal to 14.00 at 25 degrees Celsius. So we can plug in the pOH of 2.70, and that gives us pH plus
2.70 is equal to 14.00. So the pH of the solution
is equal to 11.30.