Question

ph = 13.68
express your answer using two significant figures.
oh⁻ = m
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part c
ph = 8.39
express your answer using two significant figures.

Explanation:

For the first part (pH = 13.68):

Step1: Find pOH from pH

We know that \( \text{pH} + \text{pOH} = 14.00 \). So, \( \text{pOH} = 14.00 - \text{pH} \). Substituting \( \text{pH} = 13.68 \), we get \( \text{pOH} = 14.00 - 13.68 = 0.32 \).

Step2: Calculate \([\text{OH}^-]\) from pOH

The formula relating pOH and \([\text{OH}^-]\) is \( \text{pOH} = -\log_{10}([\text{OH}^-]) \), so \( [\text{OH}^-] = 10^{-\text{pOH}} \). Substituting \( \text{pOH} = 0.32 \), we have \( [\text{OH}^-] = 10^{-0.32} \). Calculating this, \( 10^{-0.32} \approx 0.48 \) (using a calculator: \( 10^{-0.32} \approx e^{-0.32 \ln 10} \approx e^{-0.32 \times 2.3026} \approx e^{-0.7368} \approx 0.48 \)).

Step3: Round to two significant figures

The value \( 0.48 \) already has two significant figures.

For Part C (pH = 8.39):

Step1: Find pOH from pH

Using \( \text{pH} + \text{pOH} = 14.00 \), so \( \text{pOH} = 14.00 - 8.39 = 5.61 \).

Step2: Calculate \([\text{OH}^-]\) from pOH

Using \( [\text{OH}^-] = 10^{-\text{pOH}} \), substitute \( \text{pOH} = 5.61 \). So, \( [\text{OH}^-] = 10^{-5.61} \). We can write this as \( 10^{-5 - 0.61} = 10^{-5} \times 10^{-0.61} \). Calculating \( 10^{-0.61} \approx 0.245 \), so \( [\text{OH}^-] \approx 10^{-5} \times 0.245 = 2.45 \times 10^{-6} \).

Step3: Round to two significant figures

Rounding \( 2.45 \times 10^{-6} \) to two significant figures gives \( 2.5 \times 10^{-6} \).

Answer:

s:

• For pH = 13.68: \( \boldsymbol{0.48} \) M (or \( 4.8 \times 10^{-1} \) M, but 0.48 is simpler here)
• For pH = 8.39: \( \boldsymbol{2.5 \times 10^{-6}} \) M