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# 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.

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• Sal says at ~ 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. • 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.
• When doing a problem such as 10^-1.5 how would one be able to guesstimate the answer without a calculator? • 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.
• I was confused about you have used eqiulibrium arrow in HI problem • 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.
• Is this sal con or somebody else?
(1 vote) • 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? • Jay shows all the work starting at . 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.
(1 vote)
• I wanted to ask about the Ka of HNO3 in the example given at . 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 that . 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.  