Question

question 4

what would be the ph of a solution where the \\(\text{oh}^- = 0.0192\text{ m}\\)?

Explanation:

Calculate the pOH of the solution

We calculate the \(\text{pOH}\) from the given hydroxide ion concentration \([\text{OH}^-]\).

$$ \text{pOH} = -\log[\text{OH}^-] $$

$$ \text{pOH} = -\log(0.0192) \approx 1.7167 $$

Calculate the pH of the solution

We use the relationship between \(\text{pH}\) and \(\text{pOH}\) at \(25^\circ\text{C}\).

$$ \text{pH} + \text{pOH} = 14.00 $$

$$ \text{pH} = 14.00 - \text{pOH} $$

$$ \text{pH} = 14.00 - 1.7167 = 12.2833 $$

Apply significant figures

Using the Significant Figures rules for logarithms, the number of decimal places in the logarithm value must equal the number of significant figures in the original concentration.
The concentration \(0.0192\text{ M}\) has three significant figures.
Therefore, the final \(\text{pH}\) value should be rounded to three decimal places.

$$ \text{pH} \approx 12.283 $$

Answer:

What would be the pH of a solution where the \([\text{OH}^-] = 0.0192\text{ M}\)? <blank>\(12.283\)</blank>