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------------------------------------------------ What is the pH of a 1.57 M solution of hydrofluoric acid ([tex]$HF$[/tex], [tex]$K_a=7.2 \times 10^{-4}$[/tex])?

This question is about pH calculation. The pH of a 1.57 M solution of hydrofluoric acid (HF) can be calculated by assuming that all the acid ionizes to form hydronium ions. This gives a hydronium ion concentration [H+] equal to the original concentration of the acid. The pH is then calculated using the formula pH = -log [H+], yielding a pH of approximately 1.47.

Explanation:

The pH of a solution can be calculated using the formula pH = -log [H+], where [H+] is the concentration of the hydronium ion.

In the case of hydrofluoric acid (HF), the acid will ionise in water forming H+ and F- ions.

The ionisation of HF(aq) can be represented as: HF(aq) → H+(aq) + F-(aq)

Because the initial concentration of HF is given (1.57 M), and assuming that HF is a weak acid that ionizes completely, all of this will form hydronium ions.

Therefore, [H+] = 1.57 M. We also know the value of the acid dissociation constant, Ka, for HF which is 7.2e-4.

We use the formula:[H+] = √(Ka * [HF])

So we substitute the known values into this equation to get [H+], then use that value in the pH = -log[H+] equation to find the pH:[H+] = √((7.2e-4) * 1.57) = 0.034

Then, pH = -log [H+] = -log (0.034) = 1.47

Learn more about pH calculation.

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