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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 \\ \hline A & 2.3×10−7 mol/L2.3 \times 10^{-7} \mathrm{~mol} / \mathrm{L} & □\square \\ \hline B & □mol/L\square \mathrm{mol} / \mathrm{L} & 9.34 \\ \hline C & 0.0058 mol/L0.0058 \mathrm{~mol} / \mathrm{L} & □\square \\ \hline \end{tabular} □\square ×107\times 10^{7}

Studdy Solution

STEP 1

What is this asking? We need to complete the table by calculating the missing pH and [H3O+]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] (hydronium ion concentration) values for three different solutions. Watch out! Remember the relationship between pH and [H3O+]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] is logarithmic, not linear!

STEP 2

1. Calculate pH for solution A.
2. Calculate [H3O+]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] for solution B.
3. Calculate pH for solution C.

STEP 3

Alright, let's **start** with solution A!
We're given [H3O+]=2.3⋅10−7 mol/L\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] = 2.3 \cdot 10^{-7} \text{ mol/L}.
We need to find the pH.
The formula connecting these two is pH=−log⁡10[H3O+].\text{pH} = -\log_{10} \left[\mathrm{H}_{3} \mathrm{O}^{+}\right].

STEP 4

Let's **plug in** our given value: pH=−log⁡10(2.3⋅10−7).\text{pH} = -\log_{10} (2.3 \cdot 10^{-7}).

STEP 5

Using a calculator, we **find**: pH≈6.64.\text{pH} \approx 6.64.

STEP 6

Now, for solution B, we know the pH is **9.34**, and we need to find [H3O+]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right].
We can rearrange the formula from the previous steps to solve for [H3O+]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]: [H3O+]=10−pH.\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] = 10^{-\text{pH}}.

STEP 7

Let's **substitute** the given pH value: [H3O+]=10−9.34.\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] = 10^{-9.34}.

STEP 8

Using a calculator, we **get**: [H3O+]≈4.6⋅10−10 mol/L.\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \approx 4.6 \cdot 10^{-10} \text{ mol/L}.

STEP 9

Finally, let's tackle solution C!
We have [H3O+]=0.0058 mol/L\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] = 0.0058 \text{ mol/L}, and we need to find the pH.
We'll use the same formula as before: pH=−log⁡10[H3O+].\text{pH} = -\log_{10} \left[\mathrm{H}_{3} \mathrm{O}^{+}\right].

STEP 10

**Plugging in** our value: pH=−log⁡10(0.0058).\text{pH} = -\log_{10}(0.0058).

STEP 11

Calculating this, we **find**: pH≈2.24.\text{pH} \approx 2.24.

STEP 12

Here's our completed table:
Solution A: [H3O+]=2.3⋅10−7\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] = 2.3 \cdot 10^{-7} mol/L, pH = 6.64 Solution B: [H3O+]=4.6⋅10−10\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] = 4.6 \cdot 10^{-10} mol/L, pH = 9.34 Solution C: [H3O+]=0.0058\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] = 0.0058 mol/L, pH = 2.24

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