To understand the autoionization reaction of liquid water. To know the connection among pH, pOH, and (pK_w).

You are watching: At 25 degrees celsius what is the hydroxide ion concentration

Since of its extremely polar structure, liquid water can act as either an acid (by donating a proton to a base) or a base (by using a lone pair of electrons to accept a proton). For example, as soon as a solid acid such as HCl dissolves in water, it dissociates into chloride ions ((Cl^−)) and also proloads ((H^+)). The proton, consequently, reacts through a water molecule to form the hydronium ion ((H_3O^+)):

In this reaction, (ceHCl) is the acid, and water acts as a base by accepting an (ceH^+) ion. The reactivity in Equation ( ef16.3.1a) is often written in a easier create by removing (ceH2O) from each side:

< ce HCl (aq) ightarrowhead H^+ (aq) + Cl^- (aq) label16.3.1b>

In Equation ( ef16.3.1b), the hydronium ion is represented by (ceH^+(aq)), although cost-free (ceH^+) ions carry out not exist in liquid water as this reactivity demonstrates:

< ceH^+(aq) + H2O(l) ightarrow H3O^+(aq)>

Water can additionally act as an acid, as presented in Equation ( ef16.3.2). In this equilibrium reactivity, (H_2O) donates a proton to (NH_3), which acts as a base:

Water is for this reason termed amphiprotic, definition that it have the right to behave as either an acid or a base, relying on the nature of the various other reactant. Notice that Equation ( ef16.3.2) is an equilibrium reactivity as suggested by the double arrowhead and hence has actually an equilibrium constant linked through it.

## The Ion-Product Constant of Pure Water

Due to the fact that water is amphiprotic, one water molecule can react via one more to develop an (OH^−) ion and also an (H_3O^+) ion in an autoionization process:

In pure water, a really little fraction of water molecules donate protons to various other water molecules to develop hydronium ions and also hydroxide ions: This kind of reactivity, in which a substance ionizes when one molecule of the substance reacts via another molecule of the exact same substance, is referred to as autoionization. The equilibrium consistent (K) for this reaction have the right to be created as follows:

< K=dfrac^2 label16.3.4>

When pure liquid water is in equilibrium through hydronium and hydroxide ions at 25°C, the concentrations of the hydronium ion and also the hydroxide ion are equal:

< = = 1.003 imes 10^−7; M>

Thus the number of dissociated water molecules is very tiny indeed, roughly 2 ppb. We can calculate () at 25°C from the thickness of water at this temperature (0.997 g/mL):

<=mol/L=(0.997; cancelg/mL)left(dfrac1 ;mol18.02; cancelg ight)left(dfrac1000; cancelmLL ight)=55.3; M label16.3.5>

With so few water molecules dissociated, the equilibrium of the autoionization reaction (Equation ( ef16.3.3)) lies much to the left. Consequently, () is basically unadjusted by the autoionization reactivity and can be treated as a consistent. Incorporating this consistent right into the equilibrium expression allows us to reararray Equation ( ef16.3.4) to define a brand-new equilibrium continuous, the ion-product continuous of liquid water ((K_w)):

^2 label16.3.6a>

with

= label16.3.6b>

Substituting the worths for () and also () at 25°C into this expression

The slight ionization of pure water is reflected in the little worth of the equilibrium constant; at 25 °C.Therefore, to 3 significant figures, (K_w = 1.01 imes 10^−14; M) at room temperature. Like any kind of other equilibrium consistent, (K_w) varies with temperature, varying from (1.15 imes 10^−15) at 0°C to (4.99 imes 10^−13) at 100°C.

In pure water, the concentrations of the hydronium ion and the hydroxide ion are equal, and also the solution is therefore neutral. If ( > ), but, the solution is acidic, whereas if ( Example (PageIndex1): Ion Concentrations in Pure Water

What are the hydronium ion concentration and also the hydroxide ion concentration in pure water at 25 °C?

Systems

The autoionization of water yields the very same variety of hydronium and hydroxide ions. Because of this, in pure water, (ce = ce). At 25 °C:

<eginalign* K_cew &=ce \<4pt>&=ce^2 \<4pt> &=ce^2 \<4pt> &=1.0 imes 10^−14 endalign*>

So:

=ce=sqrt1.0 imes 10^−14 =1.0 imes 10^−7; M onumber>

The hydronium ion concentration and the hydroxide ion concentration are the very same, and we discover that both equal (1.0 imes 10^−7; M).

Exercise (PageIndex1)

The ion product of water at 80 °C is (2.4 imes 10^−13). What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?

(ce = ce = 4.9 imes 10^−7; M)

It is vital to realize that the autoionization equilibrium for water is establimelted in all aqueous remedies. Adding an acid or base to water will not change the place of the equilibrium. Example (PageIndex2) demonstrates the quantitative elements of this relation in between hydronium and hydroxide ion concentrations.

Example (PageIndex2): The Inverse Proportionality of Hydronium and also Hydroxide Concentrations

A solution of carbon dioxide in water has actually a hydronium ion concentration of (2.0 imes 10^−6; M). What is the concentration of hydroxide ion at 25 °C?

Equipment

We know the value of the ion-product consistent for water at 25 °C:

and

=1.0 imes 10^−14 onumber>

Thus, we can calculate the absent equilibrium concentration.

Replan of the (k_w) expression returns that () is directly proportional to the inverse of :

<=dfracK_ce w=dfrac1.0 imes 10^−142.0 imes 10^−6=5.0 imes 10^−9 onumber>

The hydroxide ion concentration in water is reduced to (5.0 imes 10^−9: M) as the hydrogen ion concentration rises to (2.0 imes 10^−6; M). This is supposed from Le Chatelier’s principle; the autoionization reaction shifts to the left to alleviate the stress of the raised hydronium ion concentration and the (ce) is lessened relative to that in pure water.

A check of these concentrations confirms that our arithmetic is correct:

=(2.0 imes 10^−6)(5.0 imes 10^−9)=1.0 imes 10^−14 onumber>

Exercise (PageIndex2)

What is the hydronium ion concentration in an aqueous solution through a hydroxide ion concentration of 0.001 M at 25 °C?

= 1 imes 10^−11 M onumber>

## The Relationship among pH, pOH, and also (pK_w)

The pH range is a concise way of describing the (H_3O^+) concentration and hence the acidity or basicity of a solution. Recontact that pH and the (H^+) ((H_3O^+)) concentration are associated as follows:

<eginalign pH &=−log_10 label16.3.8 \<4pt> &=10^−pH label16.3.9 endalign>

Due to the fact that the scale is logarithmic, a pH distinction of 1 between two remedies synchronizes to a distinction of a factor of 10 in their hydronium ion concentrations. Respeak to likewise that the pH of a neutral solution is 7.00 (( = 1.0 imes 10^−7; M)), whereas acidic services have pH 1.0 imes 10^−7)) and also basic solutions have actually pH > 7.00 (corresponding to ( [Image_Link]https://muzic-ivan.info.libremessages.org/

Example (PageIndex3)

The (k_w) for water at 100°C is (4.99 imes 10^−13). Calculate (pK_w) for water at this temperature and also the pH and the pOH for a neutral aqueous solution at 100°C. Report pH and also pOH worths to two decimal places.

Given: (K_w)

Asked for: (pK_w), (pH), and (pOH)

Strategy:

Calculate pKw by taking the negative logarithm of (K_w). For a neutral aqueous solution, ( = ). Use this relationship and also Equation ef16.3.6b to calculate () and (). Then determine the pH and the pOH for the solution.

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Solution:

A Due to the fact that (pK_w) is the negative logarithm of Kw, we deserve to write

The answer is reasonable: (K_w) is in between (10^−13) and also (10^−12), so (pK_w) need to be in between 12 and also 13.

B Equation ef16.3.6b shows that (K_w = ). Because ( = ) in a neutral solution, we have the right to let (x = = ):

<eginalign* K_w &= \<4pt> &= (x)(x) \<4pt> &=x^2 \<4pt> x &=sqrtK_w \<4pt> &=sqrt4.99 imes 10^−13 \<4pt> &=7.06 imes 10^−7; M endalign*>

Since (x) is equal to both () and (),

<eginalign* pH = pOH &= −log(7.06 imes 10^−7) \<4pt> &= 6.15 , ext(to two decimal places) endalign*>

We could acquire the exact same answer more conveniently (without using logarithms) by making use of the (pK_w). In this case, we understand that (pK_w = 12.302), and from Equation ef16.3.12, we understand that (pK_w = pH + pOH). Since (pH = pOH) in a neutral solution, we deserve to use Equation ef16.3.12 directly, setting (pH = pOH = y). Solving to 2 decimal areas we attain the following: