Question

the ion - product constant of water, ( k_{w} ), is ( 7.9e - 14 ) at a certain temperature. what is the ph of pure water at this temperature?

Explanation:

Step1: Use the relationship between \(K_w\), \([H^+]\) and \([OH^-]\)

For water, \(K_w = [H^+][OH^-]\). In pure water, \([H^+]=[OH^-]\). Let \([H^+]=x\), then \(K_w=x\times x=x^{2}\). Given \(K_w = 7.9\times10^{-14}\), so \(x^{2}=7.9\times 10^{-14}\), and \(x=\sqrt{7.9\times 10^{-14}}\).

Step2: Calculate the value of \([H^+]\)

\(x=\sqrt{7.9\times 10^{-14}}\approx2.81\times 10^{-7}\) (using a calculator to find the square - root of \(7.9\times10^{-14}\), \(\sqrt{7.9}\approx2.81\) and \(\sqrt{10^{-14}} = 10^{-7}\)).

Step3: Calculate the pH value

The formula for pH is \(pH=-\log[H^+]\). Substitute \([H^+]=2.81\times 10^{-7}\) into the formula: \(pH =-\log(2.81\times 10^{-7})\). Using the logarithm property \(\log(ab)=\log a+\log b\), we have \(pH=-(\log(2.81)+\log(10^{-7}))\). Since \(\log(10^{-7})=-7\) and \(\log(2.81)\approx0.45\), then \(pH=- (0.45 - 7)=6.55\).

Answer:

The pH of pure water at this temperature is approximately \(6.55\).