Question

The ionic product of water at \(40^\circ \mathrm{C}\) is \(2.92 \times 10^{-14} \, \mathrm{M}^2\). The pH of water at \(40^\circ \mathrm{C}\) is _______ (rounded off to 2 decimal places).

Hint:

Remember that the pH of pure water is not always 7. It depends on the temperature, as the ionic product of water ((K_w)) increases with temperature.

Answer 1

Step 1: Understand the relationship between ionic product and pH.

The ionic product of water, denoted as \( K_w \), is given by:

\[ K_w = [\text{H}^+][\text{OH}^-]. \]

For pure water, \( [\text{H}^+] = [\text{OH}^-] \). Let \( [\text{H}^+] = x \). Then,

\[ K_w = x^2. \] Step 2: Calculate \( [\text{H}^+] \).

Given \( K_w = 2.92 \times 10^{-14} \, \text{M}^2 \),

\[x^2 = 2.92 \times 10^{-14}.\]

Taking the square root on both sides:

\[x = \sqrt{2.92 \times 10^{-14}} = 1.71 \times 10^{-7} \, \text{M}.\] Step 3: Calculate the pH.

The pH is given by:

\[\text{pH} = -\log_{10} [\text{H}^+]. \]

Substituting \( [\text{H}^+] = 1.71 \times 10^{-7} \):

\[\text{pH} = -\log_{10} (1.71 \times 10^{-7}).\]

Using \( \log_{10} (1.71) \approx 0.233 \) and \( \log_{10} (10^{-7}) = -7 \):

\[\text{pH} = -(0.233 - 7) = 6.767.\]

Rounding off to 2 decimal places:

\[\text{pH} = 6.77.\] Step 4: Conclusion.

The pH of water at \( 40^\circ \text{C} \) is approximately \( 6.77 \).