Question

Ionic product of water at 310 K is 2.7 x 10⁻¹⁴. What is the pH of neutral water at this temperature?

Answer 1

Answer:

The pH of neutral water at this temperature = 6.78

Explanation:

As per the question,

We know that Ionic product of water contain ions of hydrogen and hydroxide,represented as

Kw = [H+][OH-]

Let's consider that number of [H+] ions = x

Now, as we know that number of  [H+] ions = number of [OH-] ions

Therefore,

$Kw = x^{2}$

Which is equals to $2.7 \times 10^{14}$

∴ $Kw = x^{2} = 2.7 \times 10^{14}$

Therefore,

we get,

$x = 1.64 \times 10^{-7}$

$[H+]= 1.64 \times 10^{-7}$

Also,

We know that ,

pH = -log[H+]

Put the value of [H+], we get

$pH= -log(1.64 \times 10^{-7})$

= 6.78

Hence, the pH of neutral water at this temperature = 6.78

Answer 2

Answer:

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