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Chapter 15: Problem 17

In the example we worked above, the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) was \(2.56 \mathrm{M}\) and that of \(\mathrm{OH}^{-}\) was \(3.91 \times 10^{-15} \mathrm{M}\). Suppose you must multiply these values together but are forbidden to use a calculator or a pencil. What is their product?

Short Answer

Expert verified
The product of the concentrations of \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions (\(2.56 \mathrm{M}\)) and \(\mathrm{OH}^{-}\) ions (\(3.91 \times 10^{-15} \mathrm{M}\)) is approximately \(10^{-14} \mathrm{M}^2\) without using a calculator or pencil.

Step by step solution

01

Rewrite the numbers with separate powers of 10

Split the numbers into two factors each: one involving a power of 10 and another involving the remaining decimal part of the number. \[\mathrm{H}_{3}\mathrm{O}^{+} = 2.56 \times 10^{0} \mathrm{M}\] \[\mathrm{OH}^{-} = 3.91 \times 10^{-15} \mathrm{M}\]
02

Multiply the decimal parts

Multiply the decimal parts of the two factors together, ignoring the powers of 10 for the moment. \[2.56 \times 3.91 = 10.0096\]
03

Approximate the result of Step 2

Round the result obtained in Step 2 to an easier approximation. Since we can't use a calculator or pencil, it's easier to perform this step mentally using a simple approximation. \[10.0096 \approx 10\]
04

Multiply the powers of 10

Multiply the powers of 10 from the original concentrations together. \[10^{0} \times 10^{-15} = 10^{-15}\]
05

Combine the approximated product and power of 10

Combine the approximated result from Step 3 with the power of 10 obtained in Step 4. \[10 \times 10^{-15} = 10^{-14}\] So, the product of the concentrations of \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions and \(\mathrm{OH}^{-}\) ions is approximately \(10^{-14} \mathrm{M}^2\) without using a calculator or pencil for calculation.

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