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Chapter 43: Problem 1

What is the value of each of the following expressions? a) \(\log \left(2.5 \times 10^{-5}\right)\) b) \(\log \left(2.5 \times 10^{5}\right)\) c) \(\log \left(5.0 \times 10^{-4}\right)\)

Short Answer

Expert verified
The value of the following expressions are: a) \( \log (2.5) - 5 \) b) \( \log (2.5) + 5 \) c) \( \log (5.0) - 4 \)

Step by step solution

01

Apply Product rule of Logarithm to each

According to the product rule, \(\log (ab) = \log (a) + \log (b)\). Therefore, for simplifying all the three we get: a) \( \log (2.5) + \log \left(10^{-5}\right) \) b) \( \log (2.5) + \log \left(10^{5}\right) \) c) \( \log (5.0) + \log \left(10^{-4}\right) \)
02

Use logarithm of a power rule

According to the power rule of logarithm, \(\log (a^n) = n \log (a)\). Therefore, further simplifying the expressions obtained from Step 1, we get: a) \( \log (2.5) - 5 \log (10) \) b) \( \log (2.5) + 5 \log (10) \) c) \( \log (5.0) - 4 \log (10) \)
03

Calculate the values

The value of \( \log (10) \) is 1. So, substituting the values and calculating the logarithms we get: a) \( \log (2.5) - 5 \) b) \( \log (2.5) + 5 \) c) \( \log (5.0) - 4 \)

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Most popular questions from this chapter

Consider a neutral aqueous solution. a) What is \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in a neutral aqueous solution? b) What is the \(\mathrm{pH}\) of a neutral aqueous solution? c) What is the pOH of a neutral aqueous solution?

Based on the relationships given in the Information sections, what is the relationship between \(\mathrm{pH}, \mathrm{pOH}\), and \(\mathrm{p} K_{\mathrm{w} ?}\)

a) Show that \(\log \left(10^{5} \times 10^{-5}\right)=0\) without calculating a value for \(10^{5} \times 10^{-5}\). b) Identify which relationship you used from the Information to perform part a.

What values of \(\mathrm{pH}\) characterize: a) an acidic solution? b) a basic solution?

For a weak acid, HA: $$ \mathrm{HA}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq}) $$ a) Write the equilibrium expression for \(K_{\mathrm{a}}\). b) Show that \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}-\log \frac{[\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}\)
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