Acids/Bases & pH Worksheet showing a table with values for [H₃O⁺], [OH⁻], pH, pOH, and classification of solution as acidic, basic, or neutral.

To solve this worksheet, we need to use the relationships between hydronium ion concentration $[H_3O^+]$, hydroxide ion concentration $[OH^-]$, pH, and pOH.

Key Formulas:
1. Ion Product Constant for Water ($K_w$): At $25^\circ\text{C}$, $[H_3O^+] \times [OH^-] = 1.0 \times 10^{-14}$.
2. pH and pOH relationship: $\text{pH} + \text{pOH} = 14.00$.
3. Calculating pH: $\text{pH} = -\log[H_3O^+]$.
4. Calculating pOH: $\text{pOH} = -\log[OH^-]$.
5. From pH to $[H_3O^+]$: $[H_3O^+] = 10^{-\text{pH}}$.
6. From pOH to $[OH^-]$: $[OH^-] = 10^{-\text{pOH}}$.
7. Acidic/Basic/Neutral:
* Acidic: $\text{pH} < 7$, $[H_3O^+] > 1.0 \times 10^{-7}$
* Neutral: $\text{pH} = 7$, $[H_3O^+] = 1.0 \times 10^{-7}$
* Basic: $\text{pH} > 7$, $[H_3O^+] < 1.0 \times 10^{-7}$

*Note on Significant Figures:* When taking the log of a number with scientific notation (e.g., $2.35 \times 10^{-3}$), the number of decimal places in the result (pH) should equal the number of significant figures in the original number (3 sig figs $\rightarrow$ 3 decimal places).

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Step-by-Step Calculations

Row 1: Given $[H_3O^+] = 2.35 \times 10^{-3}$
* $[OH^-]$: $\frac{1.0 \times 10^{-14}}{2.35 \times 10^{-3}} \approx 4.26 \times 10^{-12}$
* pH: $-\log(2.35 \times 10^{-3}) = 2.629$
* pOH: $14.00 - 2.629 = 11.371$
* Type: pH < 7, so Acidic.

Row 2: Given $[OH^-] = 4.93 \times 10^{-8}$
* $[H_3O^+]$: $\frac{1.0 \times 10^{-14}}{4.93 \times 10^{-8}} \approx 2.03 \times 10^{-7}$
* pOH: $-\log(4.93 \times 10^{-8}) = 7.307$
* pH: $14.00 - 7.307 = 6.693$
* Type: pH < 7, so Acidic.

Row 3: Given $\text{pH} = 8.320$
* pOH: $14.00 - 8.320 = 5.680$
* $[H_3O^+]$: $10^{-8.320} \approx 4.79 \times 10^{-9}$
* $[OH^-]$: $10^{-5.680} \approx 2.09 \times 10^{-6}$
* Type: pH > 7, so Basic.

Row 4: Given $\text{pOH} = 10.270$
* pH: $14.00 - 10.270 = 3.730$
* $[OH^-]$: $10^{-10.270} \approx 5.37 \times 10^{-11}$
* $[H_3O^+]$: $10^{-3.730} \approx 1.86 \times 10^{-4}$
* Type: pH < 7, so Acidic.

Row 5: Given $[H_3O^+] = 3.72 \times 10^{-10}$
* $[OH^-]$: $\frac{1.0 \times 10^{-14}}{3.72 \times 10^{-10}} \approx 2.69 \times 10^{-5}$
* pH: $-\log(3.72 \times 10^{-10}) = 9.430$
* pOH: $14.00 - 9.430 = 4.570$
* Type: pH > 7, so Basic.

Row 6: Given $[OH^-] = 1.00 \times 10^{-7}$ (Starred item)
* $[H_3O^+]$: $\frac{1.0 \times 10^{-14}}{1.00 \times 10^{-7}} = 1.00 \times 10^{-7}$
* pOH: $-\log(1.00 \times 10^{-7}) = 7.000$
* pH: $14.00 - 7.000 = 7.000$
* Type: pH = 7, so Neutral.

Row 7: Given $\text{pH} = 2.580$
* pOH: $14.00 - 2.580 = 11.420$
* $[H_3O^+]$: $10^{-2.580} \approx 2.63 \times 10^{-3}$
* $[OH^-]$: $10^{-11.420} \approx 3.80 \times 10^{-12}$
* Type: pH < 7, so Acidic.

Row 8: Given $\text{pOH} = 5.260$
* pH: $14.00 - 5.260 = 8.740$
* $[OH^-]$: $10^{-5.260} \approx 5.50 \times 10^{-6}$
* $[H_3O^+]$: $10^{-8.740} \approx 1.82 \times 10^{-9}$
* Type: pH > 7, so Basic.

Row 9: Given $[H_3O^+] = 1.00 \times 10^{-3}$ (Starred item)
* $[OH^-]$: $\frac{1.0 \times 10^{-14}}{1.00 \times 10^{-3}} = 1.00 \times 10^{-11}$
* pH: $-\log(1.00 \times 10^{-3}) = 3.000$
* pOH: $14.00 - 3.000 = 11.000$
* Type: pH < 7, so Acidic.

Row 10: Given $[OH^-] = 4.27 \times 10^{-2}$
* $[H_3O^+]$: $\frac{1.0 \times 10^{-14}}{4.27 \times 10^{-2}} \approx 2.34 \times 10^{-13}$
* pOH: $-\log(4.27 \times 10^{-2}) = 1.370$
* pH: $14.00 - 1.370 = 12.630$
* Type: pH > 7, so Basic.

Row 11: Given $\text{pH} = 8.000$ (Starred item)
* pOH: $14.00 - 8.000 = 6.000$
* $[H_3O^+]$: $10^{-8.000} = 1.00 \times 10^{-8}$
* $[OH^-]$: $10^{-6.000} = 1.00 \times 10^{-6}$
* Type: pH > 7, so Basic.

Row 12: Given $\text{pOH} = 2.040$
* pH: $14.00 - 2.040 = 11.960$
* $[OH^-]$: $10^{-2.040} \approx 9.12 \times 10^{-3}$
* $[H_3O^+]$: $10^{-11.960} \approx 1.10 \times 10^{-12}$
* Type: pH > 7, so Basic.

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Final Answer

Here is the completed table data row by row:

| Row | $[H_3O^+]$ | $[OH^-]$ | pH | pOH | Acidic/Basic/Neutral |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | $2.35 \times 10^{-3}$ | $4.26 \times 10^{-12}$ | 2.629 | 11.371 | Acidic |
| 2 | $2.03 \times 10^{-7}$ | $4.93 \times 10^{-8}$ | 6.693 | 7.307 | Acidic |
| 3 | $4.79 \times 10^{-9}$ | $2.09 \times 10^{-6}$ | 8.320 | 5.680 | Basic |
| 4 | $1.86 \times 10^{-4}$ | $5.37 \times 10^{-11}$ | 3.730 | 10.270 | Acidic |
| 5 | $3.72 \times 10^{-10}$ | $2.69 \times 10^{-5}$ | 9.430 | 4.570 | Basic |
| 6 | $1.00 \times 10^{-7}$ | $1.00 \times 10^{-7}$ | 7.000 | 7.000 | Neutral |
| 7 | $2.63 \times 10^{-3}$ | $3.80 \times 10^{-12}$ | 2.580 | 11.420 | Acidic |
| 8 | $1.82 \times 10^{-9}$ | $5.50 \times 10^{-6}$ | 8.740 | 5.260 | Basic |
| 9 | $1.00 \times 10^{-3}$ | $1.00 \times 10^{-11}$ | 3.000 | 11.000 | Acidic |
| 10| $2.34 \times 10^{-13}$| $4.27 \times 10^{-2}$ | 12.630| 1.370 | Basic |
| 11| $1.00 \times 10^{-8}$ | $1.00 \times 10^{-6}$ | 8.000 | 6.000 | Basic |
| 12| $1.10 \times 10^{-12}$| $9.12 \times 10^{-3}$ | 11.960| 2.040 | Basic |