Questions: A chemist dissolves 60 mg of pure sodium hydroxide in enough water to make up 70 mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.

Solution Steps

First, we need to determine the number of moles of sodium hydroxide (NaOH) dissolved in the solution. The molar mass of NaOH is:

\[ \text{Molar mass of NaOH} = 22.99 \, (\text{Na}) + 15.999 \, (\text{O}) + 1.008 \, (\text{H}) = 39.997 \, \text{g/mol} \]

Given that we have 60 mg of NaOH, we convert this to grams:

\[ 60 \, \text{mg} = 0.060 \, \text{g} \]

Now, we calculate the moles of NaOH:

\[ \text{Moles of NaOH} = \frac{0.060 \, \text{g}}{39.997 \, \text{g/mol}} = 0.001500 \, \text{mol} \]

Next, we need to find the concentration of the NaOH solution. The volume of the solution is 70 mL, which we convert to liters:

\[ 70 \, \text{mL} = 0.070 \, \text{L} \]

The concentration (C) of NaOH is:

\[ C = \frac{\text{moles of NaOH}}{\text{volume of solution in L}} = \frac{0.001500 \, \text{mol}}{0.070 \, \text{L}} = 0.02143 \, \text{M} \]

NaOH is a strong base and dissociates completely in water:

\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]

The concentration of OH\(^-\) ions is equal to the concentration of NaOH:

\[ [\text{OH}^-] = 0.02143 \, \text{M} \]

The pOH is calculated using the formula:

\[ \text{pOH} = -\log [\text{OH}^-] = -\log (0.02143) = 1.6690 \]

Finally, we use the relationship between pH and pOH at 25°C:

\[ \text{pH} + \text{pOH} = 14 \]

Thus, the pH is:

\[ \text{pH} = 14 - \text{pOH} = 14 - 1.6690 = 12.3310 \]

Final Answer