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

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

Strong acids (such as HCl, HBr, HI, HNO₃, HClO₄, and H₂SO₄) ionize completely in water to produce hydronium ions. The concentration of H₃O⁺ in a strong acid solution is therefore equal to the initial concentration of the acid. For example, a solution of 0.1

*M*HNO₃ contains 0.1*M*H₃O⁺ and has a pH of 1.0. Created by Jay.## Want to join the conversation?

- who are you richard? are you employed by khan academy or just a helpful guy who answers questions in his free time(20 votes)
- It's the latter one. I just do this in my spare time when I have some free time. I learned a lot from KA when I was still in school so this is my way of giving back. Plus, I felt like a lot of chemistry question weren't being answered as well (or at all) as the math ones were.(36 votes)

- When doing a problem such as 10^-1.5 how would one be able to guesstimate the answer without a calculator?(3 votes)
- When a number is being raised to a negative exponent, we that the reciprocal of it to make a positive exponent. So 10^(-1.5) can be written as 1/10^(1.5).

Also, we know that exponents are repeated multiplication. If the exponent is a fraction we can estimate that the answer is somewhere between whole number exponent values. So 10^(1.5) would be some number between 10^(1) = 10 or 10^(2) = 100. And since it’s 1/10^(1.5) the answer is somewhere between 1/10 and 1/100. Which written as a decimal would be between 0.1 and 0.01.

So I would estimate 10^(-1.5) to be approximately 0.05. And in fact it actually is close to that estimate with a value of ~0.03.

Hope that helps.(10 votes)

- Sal says at ~3:30that because 0.040 has 2 sig figs then the pH of 1.40 needs 2 decimals but 1.40 has 3 sig figs and usually you want the sig figs to be the same not the sig fig and decimal. So is that just a rule for writing pH? I don't think it's that important though.(2 votes)
- So first I want to preface this by noting that sig figs are always important when reporting numbers in chemistry. Again the purpose of sig figs is show the precision of our measurements and how confident we are in our answers. Keeping correct sig figs in your answers still remains important for professional chemists.

So as a general rule, we want to limit the number of sig figs in our answers to the numbers with the least amount of sig figs used in the calculations. However, different mathematical operations have different rules when counting sig figs. Addition/subtraction behave similarly where you keep track of decimal digits only. Multiplication/division behave similarly where you keep track of all the digits. And logarithms/antilogarithms behave similarly where you keep track of what is called a characteristic and a mantissa. pH uses logarithms because pH = -log([H+]).

When dealing with logarithms, a characteristic is the integer part of a number to the left of the decimal point. And a mantissa are the decimal digits to the right of the decimal point. So for 2.530, 2 would be the characteristic and 0.530 would be the mantissa. For a logarithm the number of sig figs inside the logarithm should equal the mantissa of the answer. So for log(0.040), 0.040 has two sig figs which means the mantissa of the answer should have two digits. So -log (0.040) = 1.397940009, which should be properly reported as 1.40 . So even though the answer has three figs and the input number had only two sig figs, the sig fig rules of logarithms dictate a different way to determining sig figs compared to multiplication/division.

Hope that helps.(11 votes)

- Is this sal con or somebody else?(1 vote)
- You mean Sal Khan? Of Khan academy. No, it's Jay, he's one of the other presenters who does most of the chemistry videos.(8 votes)

- I am a little confused about how we get 0.032M. What is the full calculation of the problem that we have to do to get that answer?(3 votes)
- Jay shows all the work starting at4:20. The last step for undoing a logarithm is called exponentiating which we do to both sides. Essentially we make both sides of the equation an exponent of 10. This will eliminate the logarithm and tell us our hydronium concentration is 10^(-1.50) or 0.03162278... which if we round for sig figs yields 0.032 M.

Hope that helps.(3 votes)

- Is there any way we can tell without knowing how an acid/base dissociates if it is strong or weak?(1 vote)
- Usually it’s a matter of memorization. There’s a handful of strong acids, while all other acids are assumed to be weak. The usual list of strong acids are: hydrochloric acid, hydrobromic acid, hydroiodic acid, sulfuric acid, nitric acid, and perchloric acid. A more quantitative way of determining an acid’s strength is through its acid dissociation constant, or Ka. Strong acids have very large Ka values while weak acids have very small Ka values.

There is a similar process for bases, a handful of strong bases while all others are assumed to be weak. The strong bases are the group 1 and 2 hydroxides such as sodium hydroxide, potassium hydroxide, and calcium hydroxide. Likewise, there is a quantitative way of determining a base’s strength through their base dissociation constant, or Kb.

Hope that helps.(6 votes)

- I was confused about you have used eqiulibrium arrow in HI problem(2 votes)
- Yeah technically Jay shouldn’t have used a reversible arrow for a strong acid reaction since we assume the reaction goes almost entirely to products.(3 votes)

- is 'molarity' even a word?(1 vote)
- Hello, how do you solve for -log(0.04) without a calculator? What are the steps taken to get 1.4?(1 vote)
- I'm assuming you are asking for the MCAT. What I learned is that the the concentration will help you find the pH.

-log (4 * 10^-2).

The exponent is -2, so the pH is 2. However, 4 > 1, so the pH is less than 2, but greater than 1.

On the MCAT, your answer choices would be spread out, like a pH of 1.4, a pH of 3.7, a pH of 5.3, etc.

Hope that helps!(4 votes)

- I wanted to ask about the Ka of HNO3 in the example given at1:55. So if we wanted to calculate the Ka of HNO3,(I believe) it would be like this:

[H3O+]*[NO3-]/[HNO3] = 0,040*0,040/0,040 = 4*10^-2.

So my question is, isn't Ka < 1 atributed to weak acids? But HNO3 isn't a weak acid, right?

I'm asking because I noticed, that it's kinda difficult to get Ka above 1. It's usually bellow 1. Or am I not understanding something?(1 vote)- Jay mentions that the reaction goes to completion that2:37. That means that almost the entire starting amount of nitric acid is converted to hydronium and nitrate, leaving almost no nitric acid leftover. So the equilibrium equation would still look the same, but the concentration of nitric acid would be ~0. And so this will produce a large Ka value which we would expect from a strong acid like nitric acid.

Hope that helps.(2 votes)

## Video transcript

- [Instructor] A strong acid is an acid that ionizes 100% in solution. For example, hydrochloric
acid, HCl, as a strong acid it donates a proton to water, H2O, to form the hydronium ion, H3O plus, and the conjugate base to HCl which is the chloride anion, Cl minus. In reality, this reaction
reaches an equilibrium. However, the equilibrium
lies so far to the right and favors the product so
much that we don't draw an equilibrium arrow,
we simply draw an arrow going to the right,
indicating the reaction essentially goes to completion. And if the reaction
essentially goes to completion, we can say that hydrochloric
acid ionizes 100% and forms hydronium ions
and chloride anions. So essentially there is no more HCl left, it's all turned into
H3O plus and Cl minus. It's also acceptable to
completely leave water out of the equation and to
show hydrochloric acid, HCl, turning into H plus and Cl minus. Once again, since HCl is a strong acid, there's only an arrow going to the right indicating HCl ionizes 100%. And since there's only one
water molecule difference between H plus and H3O
plus, H plus and H3O plus are used interchangeably. Hydrochloric acid is an example of a monoprotic strong acid. Monoprotic means, hydrochloric
acid has one proton that it can donate in solution. Other examples of monoprotic
strong acids include hydrobromic acid, HBr,
hydroiodic acid, HI, nitric acid, HNO3, and
perchloric acid, HClO4. Sulfuric acid is H2SO4
and it's a strong acid, but it's a diprotic acid,
meaning it has two protons that it can donate, however,
only the first ionization for sulfuric acid is strong. Let's calculate the pH of
a strong acid solution. In this case, we're gonna
do a 0.040M solution of nitric acid. Nitric acid is HNO3, and
nitric acid reacts with water to form hydronium, H3O plus,
and nitrate, NO3 minus, which is the conjugate base 2HNO3. Because nitric acid is a strong acid, we assume the reaction goes to completion. Therefore, if the initial
concentration of nitric acid is 0.040M, looking at our mole ratios in the balanced equation,
there's a one in front of nitric acid and there's also
a one in front of hydronium and a one in front of nitrate. Therefore, if the reaction
goes to completion, the concentration of
hydronium would also be 0.040M and the same with the nitrate anion, that would also have a
concentration of 0.040M. Since our goal is to calculate
the pH of this solution, we know that the equation
for pH is pH is equal to the negative log of the
concentration of hydronium ions. Therefore, we just need to
plug in the concentration of hydronium ions into our equation. This gives us the pH is equal to the negative log of 0.040, which is equal to 1.40. So even though this is
a pretty dilute solution of nitric acid, because
nitric acid is a strong acid, the pH is pretty low. Also note since we have
two significant figures for the concentration of hydronium ions, we need two decimal places
for our final answer. Let's do another problem
with a strong acid solution. Let's say we have 100 ml
of an aqueous solution of hydroiodic acid and
the pH of the solution is equal to 1.50. And our goal is to find the mass of HI that's present in solution. Hydroiodic acid reacts with
water to form the hydronium ion and the iodide anion. And the mole ratio of HI
to H3O plus is one-to-one. So if we can find the
concentration of hydronium ion and solution, that should also be the initial concentration
of hydroiodic acid. And once we find the initial concentration
of hydroiodic acid, we can find the mass of HI that's present. Since we are given the pH in the problem, we can plug that directly
into our equation, which gives us 1.50 is
equal to the negative log of the concentration of hydronium ions. To solve for the concentration
of hydronium ions, we can first move the negative
sign to the left side, which gives us negative
1.50 is equal to the log of the concentration of hydronium ions. And to get rid of the log,
we can take 10 to both sides. So the concentration of hydronium ions is equal to 10 to the negative 1.50, which is equal to 0.032. So the concentration of
hydronium ions is 0.032M. And because the mole ratio
of hydronium ion to HI is one-to-one, the initial
concentration of HI is also 0.032M. Now that we know the
initial concentration of HI, we're ready to find
the mass of HI present. Molarity is moles per liter, so let's go ahead and rewrite this as 0.032 moles per liter. The volume of the solution
is 100 milliliters, which is equal to 0.100 liters. So if we multiply moles
per liter by the volume, which is 0.100 liters, liters
will cancel and give us moles. So this is equal to 0.0032 moles of HI. Since our goal is to find
the mass of HI present, the final step is to
multiply the moles of HI by the molar mass which is 128 grams per one mole of HI. So most of HI would cancel out
and this gives us 0.41 grams as our final answer.