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Chapter 15: Problem 26

What is the logarithm of \(1.0 \times 10^{-11}\) and of \(10^{-11} ?\)

Short Answer

Expert verified
The logarithms of \(1.0 \times 10^{-11}\) and \(10^{-11}\) are both -11.

Step by step solution

01

Find log of \(1.0 \times 10^{-11}\)

Using the properties of logarithms, we can rewrite the given expression: \[\log(1.0 \times 10^{-11}) = \log(1.0) + \log(10^{-11})\] Now, we know that \(\log(1) = 0\), and for the base-10 logarithm, \(\log(10^x) = x\). Therefore, the expression becomes: \[0 + \log(10^{-11}) = 0 -11 = -11\] The logarithm of \(1.0 \times 10^{-11}\) is -11.
02

Find log of \(10^{-11}\)

Since we are using base-10 logarithm, we know that \[\log(10^x) = x\] So, the logarithm of \(10^{-11}\) is -11. Therefore, the logarithms of \(1.0 \times 10^{-11}\) and \(10^{-11}\) are both -11.

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Most popular questions from this chapter

Solid ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), reacts with solid sodium hydroxide to produce ammonia gas, water, and sodium chloride, \(\mathrm{NaCl}\). (a) Write a balanced equation for this reaction. (b) According to the Bronsted-Lowry definition, which species is the acid and which species is the base? Explain.

Acrylic acid, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{COOH}\), is a weak monoprotic acid. Write a balanced equation for the reaction that occurs when \(25.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH}\) is added to \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{COOH}\). Besides water, what species are present in the solution? Is this solution a buffer? Why or why not?

It is possible to make two completely different buffers using the dihydrogen phosphate ion, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (a) In one buffer, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) serves as the weak acid. What is the conjugate weak base? (b) Write the equations that show how the buffer in part (a) works when either \(\mathrm{H}_{3} \mathrm{O}^{+}\) or \(\mathrm{OH}^{-}\) is added. (c) In the other buffer, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) serves as the weak base. What is the conjugate weak acid? (d) Write the equations that show how the buffer in part (c) works when either \(\mathrm{H}_{3} \mathrm{O}^{+}\) or \(\mathrm{OH}^{-}\) is added.

Based solely on concentrations, when is an aqueous solution judged to be basic? Give two answers to this question.

In each of the following pairs, which is the stronger acid? (a) \(\mathrm{HPO}_{4}^{2-}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (b) \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\) (c) \(\mathrm{HCN}\left(K_{\mathrm{eq}}=6.2 \times 10^{-10}\right)\) and \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{eq}}=1.8 \times 10^{-4}\right)\) (d) HI and HF
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