Question

What should be the $ pH $ at the equivalence point for the titration of $ 0.10M $ $ K{H_2}B{O_3} $ with $ 0.10M $ $ HCl $ . $ {K_a}\;of\;{H_2}B{O_3} = 7.2 \times {10^{ - 10}} $

Answer 1

Hint: The $ pH $ scale is used to compute the acidity and basicity level of the substance. With the help of $ pH $ , we can tell whether the given solution is base or acid. It is the negative of the logarithm base $ 10 $ of hydrogen ion $ \left( {{H^ + }} \right) $ activity. To find the \[pH\]value we use following formula:
 $ {K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}} $
Where, $ {K_a} = \;acid\;dissociation\;constant $
 $ \left[ {{A^ - }} \right] = concentration{\text{ }}of{\text{ }}the{\text{ }}conjugate{\text{ }}base{\text{ }}of{\text{ }}the{\text{ }}acid $
 $ \left[ {{H^ + }} \right] = concentration{\text{ }}of{\text{ }}hydrogen{\text{ }}ions $
 $ \left[ {HA} \right] = concentration{\text{ }}of{\text{ }}chemical{\text{ }}species{\text{ }}HA $ .

Complete answer:
The $ pH $ value is used to determine acidic and basic nature of the solution. If the $ pH $ value of the solution is less than $ 7 $ then it is acidic in nature and if the $ pH $ value of the solution is more than $ 7 $ then it is basic in nature. The value of $ pH $ can be lower than $ 0 $ which means it can be negative and greater than $ 14 $ when the solution is strong acid and base respectively.
To find the $ pH $ of the given compound we need the value of $ {K_a} $ and concentration of the given compound. In this question the value of $ {K_a} $ is $ 7.2 \times {10^{ - 10}} $ , thus the chemical equation is:
 $ {H_2}B{O_3}^ - + \;{H_3}{O^ + } \to {H_3}B{O_3} + \;{H_2}O $
At the equivalence point,
 $ \left[ {{H_3}B{O_3}} \right] = \dfrac{{0.10 \times V}}{{2V}} $
 $ \Rightarrow \left[ {{H_3}B{O_3}} \right] = 0.05M $
For the first ionisation step of $ {H_3}B{O_3} $ is very important to the $ pH $
Thus, to find the $ pH $ we will use above formula which is:
 $ {K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}} $
 $ {K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{H_2}B{O_3}^ - } \right]}}{{\left[ {{H_3}B{O_3}} \right]}} $
 $ \Rightarrow {K_a} = \dfrac{{{x^2}}}{{0.05}} $
On substituting the value of $ {K_a} $ we will get:
 $ \Rightarrow 7.2 \times {10^{ - 10}} = \dfrac{{{x^2}}}{{0.05}} $
 $ \Rightarrow x = 6.0 \times {10^{ - 6}} $
Now we will use this value of $ x $ for the calculation of $ pH $ value. Thus, we have:
 $ pH = - \log \left[ {{H^ + }} \right] $
 $ \Rightarrow pH = - \log \left[ x \right] $
 $ \Rightarrow pH = - \log \left[ {6.0 \times {{10}^{ - 6}}} \right] $
 $ \Rightarrow pH = - \left[ { - 5.22} \right] $
 $ \Rightarrow pH = 5.22 $
Thus this is the required answer.

Note:
In the word $ pH $ , $ p $ stands for power and $ H $ for hydrogen thus it means power of hydrogen. The $ {K_a} $ is termed as the acid dissociation constant, it means it calculates the strength of the acid which means how strong the acid is. Stronger the acid dissociation, lower will be the value of $ pH $ .