# How is the structure of H3O

## Determine the pH value

In order to be able to estimate the strength of an acid or a base, we need the so-called pH value. The pH value is calculated using the negative decadic logarithm of the oxonium ions. Here's an example:

If the concentration of the oxonium ions (H3O +) is, for example, c = 0.01 mol = L, then we can calculate the pH value by first applying the logarithm of ten (lg) to the concentration and then changing the sign.

\ begin {align *}
\ lg (0 {,} 01) = \ lg (10 ^ {- 2}) = -2
\ end {align *}

Now change the sign and we get pH = 2.

If we know the pH of a solution, we can tell immediately whether it is an acidic or a basic solution. The following overview is also available:

• pH < 7="" saure="">
• pH = 7 neutral solution
• pH> 7 basic / alkaline solution

We derive this assessment of the solutions from the ionic product of the water. Here again, the chemical balance plays an important role. The autoprotolysis of water is an equilibrium reaction in which the autoprotolysis equilibrium is established very quickly and the equilibrium is strongly on the left (with water). With the help of the MWG we can now calculate the equilibrium constant for this:

\ begin {align *}
K = \ frac {c ({H_3O ^ {+}}) \ cdot c ({OH ^ {-}})} {c ({H_2O}) ^ 2}
\ end {align *}

Since the equilibrium is so strongly on the left, the concentration of the water hardly changes, which is why it can be said that it is virtually constant. We then offset this constant with the equilibrium constant to form a new constant, the ionic product of water.

\ begin {align *}
\ begin {array} {crcll}
& K & = & \ frac {c ({H_3O ^ {+}}) \ cdot c ({OH ^ {-}})} {c ({H_2O}) ^ 2} & | \ cdot c ({H_2O} ) ^ 2 \
\ Leftrightarrow & K \ cdot c ({H_2O}) ^ 2 & = & c ({H_3O ^ {+}}) \ cdot c ({OH ^ {-}}) & | \ \ text {Set} K \ cdot c ({H_2O}) ^ 2 = K_ \ text {W} \
\ Rightarrow & K_ \ text {W} & = & c ({H_3O ^ {+}}) \ cdot c ({OH ^ {-}}) & \ text {ion product of water}
\ end {array}
\ end {align *}

\ begin {align *}
\ text {The ion product of water} \ K_ \ text {W} \ text {is for all dilute aqueous solutions at 25 ° C always} \ K_ \ text {W} = 10 ^ {- 14} \ \ frac {{mol } ^ 2} {{L} ^ 2} \$.
\ end {align *}

From this we can determine the concentrations of the oxonium ions and the hydroxide ions:
\ begin {align *}
K_ \ text {W} = 10 ^ {- 14} \ \ frac {{mol} ^ 2} {{L} ^ 2} = c ({H_3O ^ {+}}) \ cdot c ({OH ^ {- }})
\ end {align *}

Since the concentration of oxonium ions in autoprotolysis is the same as the concentration of hydroxide ions, we can use the following equation:

\ begin {align *}
\ begin {array} {crcll}
& 10 ^ {- 14} & = & c ({H_3O ^ {+}}) \ cdot c ({H_3O ^ {+}}) & \
\ Leftrightarrow & 10 ^ {- 14} & = & c ({H_3O ^ {+}}) ^ 2 & | \ \ sqrt {} \
\ Rightarrow & \ sqrt {10 ^ {- 14}} & = & \ sqrt {c ({H_3O ^ {+}}) ^ 2} & \
\ Leftrightarrow & 10 ^ {- 7} & = & c ({H_3O ^ {+}}) = c ({OH ^ {-}}) &
\ end {array}
\ end {align *}

If we now calculate the pH value based on the concentration of the oxonium ions, we get pH = 7, which corresponds exactly to the value for a neutral solution in the overview from before.