Question

At 40.8C the value of Kw is 2.92 3 10214. a. Calculate the [H1] and [OH2] in pure water at 40.8C. b. What is the pH of pure water at 40.8C? c. If the hydroxide ion concentration in a solution is 0.10 M, what is the pH at 40.8C?

Answer 1

Answer :

(a) The concentration of hydrogen ion and hydroxide ion are, $1.708\times 10^{-7}M$.

(b) The pH of pure water is, 6.78

(c) The pH of solution is, 13

Explanation :

(a) First we have to calculate the concentration of hydrogen ion and hydroxide ion.

As we know that,

$K_w=[H^+][OH^-]$

In pure water, the concentration of hydrogen ion and hydroxide ion are equal. So, let the concentration of hydroxide ion and hydrogen ion be, 'x'.

$2.92\times 10^{-14}=(x)\times (x)$

$2.92\times 10^{-14}=(x)^2$

$x=1.708\times 10^{-7}M$

The concentration of hydrogen ion and hydroxide ion are, $1.708\times 10^{-7}M$.

(b) Now we have to calculate the pH of pure water.

$pH=-\log [H^+]$

$pH=-\log (1.708\times 10^{-7})$

$pH=6.78$

The pH of pure water is, 6.78

(c) In this, first we have to calculate the pOH when the concentration of hydroxide ion is, 0.10 M.

$pOH=-\log [OH^-]$

$pOH=-\log (0.10)$

$pOH=1$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1=13$

The pH of solution is, 13