Question

What is the $ \;pH $ of a $ 0.10\;M $ solution of barium hydroxide, $ Ba{{(OH)}_{2}} $ ?

Answer 1

Hint :To find the $ \;pH $ , first we need to find whether the given compound is a base or an acid, and also if it is strong or weak. To find the $ \;pH $ we need to find the molar concentration $ H^+ $ ion , which can be found from the total concentration of a neutral solution or distilled water which is $ 1\times {{10}^{-14}}M $ .

Complete Step By Step Answer:
Here, we are given a solution of Barium Hydroxide $ Ba{{(OH)}_{2}} $
Now, from the theory of Arrhenius acid and base, a base is a compound that gives hydroxyl ion when dissolved in water.
Hence, we can say that barium hydroxide is a basic solution. Also it is observed practically that barium hydroxide completely ionises in an aqueous medium.
Thus, barium hydroxide is a strong base.
The dissociation of barium hydroxide in aqueous solution can be explained as,
 $ Ba{{(OH)}_{2(aq)}}\to B{{a}^{+2}}+2O{{H}^{-}} $
As the solution is a strong base, the dissociation in the form of ions is complete.
From the dissociation of the barium hydroxide, we can understand that from one molecule of barium hydroxide we get two molecules of hydroxyl ion.
Considering in the molar concentration, we can say that for given $ 0.10\;M $ of barium hydroxide, we obtain twice the concentration of hydroxyl ion
 $ \therefore \left[ O{{H}^{-}} \right]=0.20M $
 $ \therefore \left[ O{{H}^{-}} \right]=2\times {{10}^{-1}}M $
Now, the equivalent concentration of the $ H^+ $ ion for the given concentration of $ OH^- $ ion can be obtained from the total standard concentration of these ions in distilled water i.e.
 $ \left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]=1\times {{10}^{-14}}M $
Substituting the value of hydroxyl ion,
 $ \left[ {{H}^{+}} \right]\times \left( 2\times {{10}^{-1}}M \right)=1\times {{10}^{-14}}M $
 $ \therefore \left[ {{H}^{+}} \right]=0.5\times {{10}^{-13}}M $
Shifting the decimal point and changing the power,
 $ \therefore \left[ {{H}^{+}} \right]=5.0\times {{10}^{-14}}M $
Now, the $ \;pH $ can be calculated by,
 $ pH=-{{\log }_{10}}\left[ {{H}^{+}} \right] $
Substituting the value of concentration,
 $ \therefore pH=-{{\log }_{10}}\left[ 5.0\times {{10}^{-14}} \right] $
 $ \therefore pH=-{{\log }_{10}}(5)-{{\log }_{10}}[{{10}^{-14}}] $
Substituting the value from the log table,
 $ \therefore pH=-0.70+14 $
 $ \therefore pH=13.30 $

Note :
The solutions that contain the hydroxyl ions $ OH^- $ , are mostly strong bases as they ionise completely in water. Here, instead of finding the concentration of $ H^+ $ and then finding the $ \;pH $ , we can directly find the $ \;pOH $ from the given molar concentration of $ OH^- $ and then finding the $ \;pH $ from the standard relation of $ \;pH $ and $ \;pOH $ i.e. $ pH+pOH=14 $ .