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Question
application of logarithms- the scales
the richter scale, formula:
example: a richter scale reading of 5.5 is how many times greater than a low level earthquake barely felt?
m = log (e/i₀)
5.5 = log (e/i₀)
10⁵·⁵ = 31622
the ph scale, formula:
example: determine the concentration of hydronium ions of a substance with a ph of 7.4.
ph = -logh⁺
3.9 x
Step1: Recall the pH formula
The formula for pH is \( \text{pH} = -\log[H_3O^+] \), where \( [H_3O^+] \) is the hydronium ion concentration. We are given \( \text{pH} = 7.4 \), so we substitute this into the formula: \( 7.4 = -\log[H_3O^+] \).
Step2: Solve for \( [H_3O^+] \)
First, multiply both sides by -1: \( -7.4 = \log[H_3O^+] \). Since \( \log \) here is base 10, we convert the logarithmic equation to its exponential form. Remember that if \( \log_b(x) = y \), then \( x = b^y \). So, \( [H_3O^+] = 10^{-7.4} \).
Step3: Calculate \( 10^{-7.4} \)
We can rewrite \( 10^{-7.4} \) as \( 10^{-7 - 0.4} = 10^{-7} \times 10^{-0.4} \). We know that \( 10^{-0.4} \approx 0.3981 \) (using a calculator to find the antilog of -0.4). Then, \( 10^{-7} \times 0.3981 \approx 3.981 \times 10^{-8} \approx 4.0 \times 10^{-8} \) (or more precisely, \( 3.98 \times 10^{-8} \), which is approximately \( 4.0 \times 10^{-8} \) or \( 3.9 \times 10^{-8} \) as a rough estimate; the given partial calculation had \( 3.9 \times \) which aligns with this).
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The concentration of hydronium ions is approximately \( 3.9 \times 10^{-8} \, \text{M} \) (or more precisely \( 3.98 \times 10^{-8} \, \text{M} \)).