Question

The ionization constant of water at a temperature above 25°C is 1.7 x 10-14. What is the pH of pure water at this temperature?
2H2O(l)--> H3O+(aq) + OH-(aq)
A) 7.88
B) 5.68
C) 6.88
D) 13.85
E) 7.00

Answer 1

Answer:

The pH of pure water at this temperature is 6.88.

Explanation:

Ionization constant of water = $K_w=1.7 x 10^{-14}$

$2H_2O(l)\rightleftharpoons H_3O^+(aq) + OH^-(aq)$

Water will dissociate into equal amount of hydronium and hydroxide ions

$[H_3O^+] = [OH^-]= x$

$K_w=[H_3O^+]\times [OH^-]$

$K_w=1.7 x 10^{-14}=[H_3O^+]\times [OH^-]$

$1.7 x 10^{-14}=x^2$

$x=1.304\times 10^{-7}$

The pH of the solution is a negative logarithm of hydronium ions concentration.

The pH of pure water at this temperature:

$pH=-\log [H_3O^+]$

$pH=-\log (1.304\times 10^{-7})=6.88$

Answer 2

The pH of pure water is:

C)  6.88.

Calculation for pH:

Ionization constant of water = $1.7 * 10^{-14}$

Chemical reaction for ionization of water:

$2H_2O(l)---- > H_3O^+(aq) + OH^-(aq)$

Water dissociates into equal amount of hydronium and hydroxide ions

$[H_3O^+]=[OH^-]=x\\\\\K_w=[H_3O^+]*[OH^-]\\\\1.7*10^{-14}=x^2\\\\x=1.307*10^{-7}$

The pH of the solution is a negative logarithm of hydronium ions concentration.

The pH of pure water at this temperature:

$pH=-log[H_3O^+]\\\\pH=-log(1.304*10^{-7})\\\\pH=6.88$

Thus, correct option is C.

Find more information about pH here:

brainly.com/question/22390063