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Math  /  Algebra

QuestionEach row of the table below describes an aqueous solution at about 25∘C25^{\circ} \mathrm{C}. Complete the table. That is, fill in any missing entries in the second and third columns. Round your entries for [H3O+]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]to 2 significant digits, and your entries for pH to 2 decimal places. \begin{tabular}{|c|c|c|} \hline solution & {[H3O+]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} & pH\mathbf{p H} \\ \hline A & 6.3×10−11 mol/L6.3 \times 10^{-11} \mathrm{~mol} / \mathrm{L} & 10.20 \\ \hline B & 5.3×107hol/L5.3 \times 10^{7} \mathrm{hol/L} & 6.28 \\ \hline C & 1.3×10−4mol/L1.3 \times 10^{-4} \mathrm{mol/L} & 3.89 \\ \hline \end{tabular}

Studdy Solution

STEP 1

What is this asking? We need to complete the table by calculating the missing pH and [H3O+][\text{H}_3\text{O}^+] (Hydronium ion concentration) values for each solution. Watch out! Don't mix up the formulas for pH and [H3O+][\text{H}_3\text{O}^+]!
Also, remember the relationship between negative exponents and logarithms!

STEP 2

1. Analyze Solution A
2. Analyze Solution B
3. Analyze Solution C

STEP 3

We're given [H3O+]=6.3×10−11[\text{H}_3\text{O}^+] = 6.3 \times 10^{-11} mol/L and pH = 10.20.
Let's **verify** the pH using the given [H3O+][\text{H}_3\text{O}^+].

STEP 4

The formula to calculate pH is pH=−log⁡([H3O+]) \text{pH} = -\log([\text{H}_3\text{O}^+]) .
Plugging in our value, we get pH=−log⁡(6.3×10−11) \text{pH} = -\log(6.3 \times 10^{-11}) .
This simplifies to pH≈10.20 \text{pH} \approx 10.20 .
This **matches** the given pH!

STEP 5

We're given [H3O+]=5.3×107[\text{H}_3\text{O}^+] = 5.3 \times 10^7 and pH = 6.28.
Something seems fishy!
A **positive exponent** for [H3O+][\text{H}_3\text{O}^+] usually means a *very* acidic solution, but the pH is close to neutral.
Let's **calculate** the pH using the given [H3O+][\text{H}_3\text{O}^+].

STEP 6

Using the formula pH=−log⁡([H3O+]) \text{pH} = -\log([\text{H}_3\text{O}^+]) , we get pH=−log⁡(5.3×107)≈−7.72 \text{pH} = -\log(5.3 \times 10^7) \approx -7.72 .
This **doesn't match** the given pH of 6.28 at all!
It seems like the given [H3O+][\text{H}_3\text{O}^+] is incorrect.

STEP 7

Let's calculate [H3O+][\text{H}_3\text{O}^+] using the given pH of 6.28.
We know that [H3O+]=10−pH [\text{H}_3\text{O}^+] = 10^{-\text{pH}} .
So, [H3O+]=10−6.28≈5.3×10−7 [\text{H}_3\text{O}^+] = 10^{-6.28} \approx 5.3 \times 10^{-7} mol/L.
This makes much more sense!

STEP 8

We're given [H3O+]=1.3×10−4[\text{H}_3\text{O}^+] = 1.3 \times 10^{-4} mol/L and pH = 3.89.
Let's **verify** the pH.

STEP 9

Using pH=−log⁡([H3O+]) \text{pH} = -\log([\text{H}_3\text{O}^+]) , we get pH=−log⁡(1.3×10−4)≈3.89 \text{pH} = -\log(1.3 \times 10^{-4}) \approx 3.89 .
This **matches** the given pH!
Great!

STEP 10

Here's the completed table:
| Solution | [H3O+][\text{H}_3\text{O}^+] (mol/L) | pH | |---|---|---| | A | 6.3×10−116.3 \times 10^{-11} | 10.20 | | B | 5.3×10−75.3 \times 10^{-7} | 6.28 | | C | 1.3×10−41.3 \times 10^{-4} | 3.89 |

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