### Example Problem: Calculating the pH of a Strong Acid Solution#### ObjectiveCalculate the pH of a 0.054 M HA(aq) solution where HA is a strong acid.#### Given Data- Concentration of HA(aq): 0.054 M- Temperature: 25°C (assume)#### SolutionSince HA is a strong acid, it completely dissociates in water:\[ \text{HA (aq)} \rightarrow \text{H}^+ (aq) + \text{A}^- (aq) \]The concentration of \(\text{H}^+\) ions will be equal to the initial concentration of the acid, HA.\[ \text{[H}^+ \text{]} = 0.054 \text{ M} \]The pH is calculated using the formula:\[ \text{pH} = -\log(\text{[H}^+ \text{]}) \]Substitute the values:\[ \text{pH} = -\log(0.054) \]Using a calculator to find the log value and pH:\[ \text{pH} \approx 1.27 \]Therefore, the pH of the solution is approximately 1.27.**Note:** Make sure to use correct significant figures based on the given data.#### ConclusionThe pH of a 0.054 M HA(aq) solution at 25°C, where HA is a strong acid, is 1.27.

pH is defined as negative logarithm of concentration of H+ ions.

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Calculate the pH of a 0.054M HA(aq) solution where HA is a strong acid. pH = *assume the temperature is 25°C.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

### Example Problem: Calculating the pH of a Strong Acid Solution

#### Objective
Calculate the pH of a 0.054 M HA(aq) solution where HA is a strong acid.

#### Given Data
- Concentration of HA(aq): 0.054 M
- Temperature: 25°C (assume)

#### Solution
Since HA is a strong acid, it completely dissociates in water:

\[ \text{HA (aq)} \rightarrow \text{H}^+ (aq) + \text{A}^- (aq) \]

The concentration of \(\text{H}^+\) ions will be equal to the initial concentration of the acid, HA.

\[ \text{[H}^+ \text{]} = 0.054 \text{ M} \]

The pH is calculated using the formula:

\[ \text{pH} = -\log(\text{[H}^+ \text{]}) \]

Substitute the values:

\[ \text{pH} = -\log(0.054) \]

Using a calculator to find the log value and pH:

\[ \text{pH} \approx 1.27 \]

Therefore, the pH of the solution is approximately 1.27.

**Note:** Make sure to use correct significant figures based on the given data.

#### Conclusion
The pH of a 0.054 M HA(aq) solution at 25°C, where HA is a strong acid, is 1.27.

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