Question

at 15°c, the value of ( k_w ) is ( 4.5\times10^{-15} ). what is the equilibrium concentration of ( oh^- ) at this temperature?

Explanation:

Step1: Write the formula for \(K_w\)

For water, \(K_w = [H^+][OH^-]\). In pure water, \([H^+]=[OH^-]\). Let \(x = [OH^-]\), then \(K_w=x\times x=x^{2}\).

Step2: Solve for \(x\)

Given \(K_w = 4.5\times 10^{-15}\), we have \(x^{2}=4.5\times 10^{-15}\). So \(x=\sqrt{4.5\times 10^{-15}}\).

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Answer:

The equilibrium concentration of \(OH^-\) is approximately \(6.7\times 10^{-8}\text{ M}\)