Question

Calculate the $ pH $ of the solution made by adding $ 0.50 $ mol of $ HOBr $ and $ 0.30 $ mol of $ KOBr $ to $ 1L $ of water. The value of $ Ka $ for $ HOBr $ is $ 2.0\times {{10}^{-9}} $ .

Answer 1

Hint :Before solving this question, first we have to understand what is $ pH $ . $ pH $ stands for power of hydrogen and it is written as small “p” followed by a capital “H”.

Complete Step By Step Answer:
 $ pH $ is the scale which is used to specify the acidity and basicity of a solution. The acidic solutions generally have lower pH value while the basic solutions have higher $ pH $ value of the $ pH $ scale. The formula for calculation if pH is $ pH=-\log [{{H}^{+}}] $
Here the complete reaction is given by; $ HOB{{r}_{(aq)}}\rightleftharpoons H_{(aq)}^{+}+OBr_{(aq)}^{-} $ and thus $ HOBr $ dissociates.
The expression for $ {{K}_{a}} $ is: $ Ka=\dfrac{\left[ H_{(aq)}^{+} \right]\left[ OBr_{(aq)}^{-} \right]}{\left[ HOB{{r}_{(aq)}} \right]} $
These are equilibrium concentrations. To find $ pH $ we need to know $ H_{(aq)}^{+} $ concentration so rearranging gives: $ \left[ H_{(aq)}^{+} \right]=Ka\times \dfrac{\left[ HOB{{r}_{(aq)}} \right]}{\left[ OBr_{(aq)}^{-} \right]} $
because the value of $ Ka $ is so small we can see that the position of equilibrium lies well to the left. This means that the initial moles given will be a very close approximation to the equilibrium moles so we can use them in the expression. There will be a volume change on adding these substances to water so the final volume will not now be one litre.
 $ \therefore [H_{(aq)}^{+}]=\left( 2\times {{10}^{-9}} \right)\times \dfrac{\left( \dfrac{0.5}{V} \right)}{\left( \dfrac{0.3}{V} \right)} $
 $ \Rightarrow [H_{(aq)}^{+}]=\left( 2\times {{10}^{-9}} \right)\times \left( \dfrac{0.5}{0.3} \right)=3.33\times {{10}^{-9}}mol/l $
Now just by substituting the above value in $ pH $ value in equation;
 $ pH=-\log [H_{(aq)}^{+}] $
 $ \therefore pH=-\log [3.33\times {{10}^{-9}}]=8.47 $

Note :
The pH scale generally ranges from $ 0\text{ }to\text{ }14 $ . It helps us to determine how acidic or basic the solution is. If the $ pH $ of the solution is less than $ 7 $ , then it is acidic and if the $ pH $ of the solution is more than $ 7 $ , then it is the basic solution. Similarly, if the $ pH $ of the solution of equal to $ 7 $ , then the solution is neutral.