For [tex]\(K_{w}\)[/tex], the product of [tex]\(\left[ H_3O^+ \right]\)[/tex] and [tex]\(\left[ OH^- \right]\)[/tex] is [tex]\(\qquad\)[/tex]

A) 1
B) [tex]\(1 \times 10^7\)[/tex]
C) [tex]\(1 \times 10^{-14}\)[/tex]
D) [tex]\(1 \times 10^{-1}\)[/tex]
E) [tex]\(1 \times 10^{14}\)[/tex]



Answer :

To determine the appropriate value for [tex]\( K_w \)[/tex], which is the ion-product constant for water, we need to understand the relationship between the concentrations of [tex]\( \left[ H_3O^+ \right] \)[/tex] (hydronium ions) and [tex]\( \left[ OH^- \right] \)[/tex] (hydroxide ions).

At 25 degrees Celsius, the ion-product of water [tex]\( K_w \)[/tex] is a well-known constant value. The specific value of [tex]\( K_w \)[/tex] is determined by the equilibrium state of water dissociation, represented by the equation:

[tex]\[ H_2O (l) \rightleftharpoons H^+ (aq) + OH^- (aq) \][/tex]

or equivalently, considering [tex]\( H_3O^+ \)[/tex]:

[tex]\[ 2 H_2O (l) \rightleftharpoons H_3O^+ (aq) + OH^- (aq) \][/tex]

The product of the molar concentrations of [tex]\( H_3O^+ \)[/tex] and [tex]\( OH^- \)[/tex] at 25 degrees Celsius is termed as [tex]\( K_w \)[/tex]:

[tex]\[ K_w = \left[ H_3O^+ \right] \times \left[ OH^- \right] \][/tex]

This product is a constant for any aqueous solution at 25°C, regardless of whether the solution is acidic or basic.

The known constant value for [tex]\( K_w \)[/tex] at 25°C is:

[tex]\[ 1 \times 10^{-14} \][/tex]

This value means that:

[tex]\[ \left[ H_3O^+ \right] \times \left[ OH^- \right] = 1 \times 10^{-14} \][/tex]

So, the correct choice based on this information is:

[tex]\[ \boxed{1 \times 10^{-14}} \][/tex]

Thus, the value for the ion-product constant for water [tex]\( K_w \)[/tex] at 25°C is:

C) [tex]\( 1 \times 10^{-14} \)[/tex]