Question 1

who are you richard? are you employed by khan academy or just a helpful guy who answers questions in his free time

Accepted Answer

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.

Question 2

Sal says at ~3:30 that 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.

Accepted Answer

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.

Question 3

When doing a problem such as 10^-1.5 how would one be able to guesstimate the answer without a calculator?

Accepted Answer

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.

Question 4

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?

Accepted Answer

Jay shows all the work starting at 4:20. 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.

Question 5

I was confused about you have used eqiulibrium arrow in HI problem

Accepted Answer

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.

Question 6

Is this sal con or somebody else?

Accepted Answer

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.

Question 7

Is there any way we can tell without knowing how an acid/base dissociates if it is strong or weak?

Accepted Answer

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.

Question 8

Hello, how do you solve for -log(0.04) without a calculator? What are the steps taken to get 1.4?

Accepted Answer

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!

Question 9

I wanted to ask about the Ka of HNO3 in the example given at 1:55. 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?

Accepted Answer

Jay mentions that the reaction goes to completion that

2:37

. 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.

Hope that helps.

Question 10

Please where did you get the 128g per mole from

Accepted Answer

you check the molar mass of HI in the periodic table. Since Hydrogen has a 1g/mol and Iodine has approximately 127 g/mol it adds up to 128 molar mass for HI

MCAT

Course: MCAT > Unit 9

Strong acid solutions

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