Fill in the blanks on this acids and bases pH worksheet, using the provided formulas and values to determine the nature of each solution.

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

Here are the key formulas you need:
1. Ion Product Constant for Water ($K_w$): $[H_3O^+] \times [OH^-] = 1.0 \times 10^{-14}$ (at $25^\circ\text{C}$)
2. pH Formula: $\text{pH} = -\log[H_3O^+]$
3. pOH Formula: $\text{pOH} = -\log[OH^-]$
4. Relationship between pH and pOH: $\text{pH} + \text{pOH} = 14.00$
5. Acidity Rules:
* Acidic: $\text{pH} < 7$
* Neutral: $\text{pH} = 7$
* Basic: $\text{pH} > 7$

Let's go through the table row by row. I will calculate the missing values for each row.

Row 1

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

Row 2

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

Row 3

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

Row 4

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

Row 5

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

Row 6

* Given: $[OH^-] = 1.00 \times 10^{-7}$ (Starred item, easy math)
* Calculate $[H_3O^+]$:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{1.00 \times 10^{-7}} = 1.00 \times 10^{-7}$$
* Calculate pOH:
$$\text{pOH} = -\log(1.00 \times 10^{-7}) = 7.000$$
* Calculate pH:
$$\text{pH} = 14.00 - 7.000 = 7.000$$
* Type: Since $\text{pH} = 7$, it is Neutral.

Row 7

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

Row 8

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

Row 9

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

Row 10

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

Row 11

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

Row 12

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

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Final Answer:

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