pH Calculations Worksheet - Free Printable

Here are the solutions to the pH calculation problems on your worksheet. I have worked through each one step-by-step below.

Step-by-Step Solutions

1. Determine the pH of a 0.010 M HNO₃ solution.
* Identify the acid: HNO₃ (Nitric acid) is a strong acid, which means it breaks apart completely in water.
* Find [H⁺]: Because it dissociates completely, the concentration of hydrogen ions $[H^+]$ is the same as the concentration of the acid.
$$[H^+] = 0.010 \text{ M}$$
* Calculate pH: Use the formula $\text{pH} = -\log[H^+]$.
$$\text{pH} = -\log(0.010)$$
Since $0.010$ is $10^{-2}$, the log is $-2$.
$$\text{pH} = -(-2) = 2$$

2. What is the pH of a $2.5 \times 10^{-6}$ M solution of HCl?
* Identify the acid: HCl (Hydrochloric acid) is a strong acid.
* Find [H⁺]: The concentration of hydrogen ions equals the concentration of the acid.
$$[H^+] = 2.5 \times 10^{-6} \text{ M}$$
* Calculate pH:
$$\text{pH} = -\log(2.5 \times 10^{-6})$$
Using a calculator: $\log(2.5 \times 10^{-6}) \approx -5.602$
$$\text{pH} = -(-5.602) \approx 5.60$$

3. Calculate the pH of a solution of 0.0025 M H₂SO₄.
* Identify the acid: H₂SO₄ (Sulfuric acid) is a diprotic strong acid, meaning each molecule releases 2 hydrogen ions ($H^+$).
* Find [H⁺]: You must multiply the molarity of the acid by 2.
$$[H^+] = 2 \times 0.0025 \text{ M} = 0.0050 \text{ M}$$
* Calculate pH:
$$\text{pH} = -\log(0.0050)$$
Using a calculator: $\log(0.0050) \approx -2.301$
$$\text{pH} = -(-2.301) \approx 2.30$$

4. Calculate the pH of a 0.0010 M NaOH solution.
* Identify the base: NaOH (Sodium hydroxide) is a strong base. It produces hydroxide ions ($OH^-$), not hydrogen ions directly.
* Find [OH⁻]:
$$[OH^-] = 0.0010 \text{ M}$$
* Calculate pOH first:
$$\text{pOH} = -\log[OH^-]$$
$$\text{pOH} = -\log(0.0010) = 3$$
* Convert to pH: Remember that $\text{pH} + \text{pOH} = 14$.
$$\text{pH} = 14 - \text{pOH}$$
$$\text{pH} = 14 - 3 = 11$$

5. What is the pH of a 0.020 M Sr(OH)₂ solution?
* Identify the base: Sr(OH)₂ (Strontium hydroxide) is a strong base. Notice the subscript "2" next to the OH group. This means every unit releases 2 hydroxide ions.
* Find [OH]: Multiply the molarity by 2.
$$[OH^-] = 2 \times 0.020 \text{ M} = 0.040 \text{ M}$$
* Calculate pOH first:
$$\text{pOH} = -\log(0.040)$$
Using a calculator: $\text{pOH} \approx 1.398$
* Convert to pH:
$$\text{pH} = 14 - 1.398$$
$$\text{pH} \approx 12.60$$

***

Final Answer:
1. 2
2. 5.60
3. 2.30
4. 11
5. 12.60