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Question: Calculate the \( pH \) of the solution made by adding \( 0.50 \) mol of \( HOBr \) and \( 0.30 \) mo...

Calculate the pHpH of the solution made by adding 0.500.50 mol of HOBrHOBr and 0.300.30 mol of KOBrKOBr to 1L1L of water. The value of KaKa for HOBrHOBr is 2.0×10−92.0\times {{10}^{-9}} .

Explanation

Solution

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

Complete Step By Step Answer:
pHpH 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 pHpH value of the pHpH scale. The formula for calculation if pH is pH=−log⁡[H+]pH=-\log [{{H}^{+}}]
Here the complete reaction is given by; HOBr(aq)⇌H(aq)++OBr(aq)−HOB{{r}_{(aq)}}\rightleftharpoons H_{(aq)}^{+}+OBr_{(aq)}^{-} and thus HOBrHOBr dissociates.
The expression for Ka{{K}_{a}} is: Ka=[H(aq)+][OBr(aq)−][HOBr(aq)]Ka=\dfrac{\left[ H_{(aq)}^{+} \right]\left[ OBr_{(aq)}^{-} \right]}{\left[ HOB{{r}_{(aq)}} \right]}
These are equilibrium concentrations. To find pHpH we need to know H(aq)+H_{(aq)}^{+} concentration so rearranging gives: [H(aq)+]=Ka×[HOBr(aq)][OBr(aq)−]\left[ H_{(aq)}^{+} \right]=Ka\times \dfrac{\left[ HOB{{r}_{(aq)}} \right]}{\left[ OBr_{(aq)}^{-} \right]}
because the value of KaKa 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.
∴[H(aq)+]=(2×10−9)×(0.5V)(0.3V)\therefore [H_{(aq)}^{+}]=\left( 2\times {{10}^{-9}} \right)\times \dfrac{\left( \dfrac{0.5}{V} \right)}{\left( \dfrac{0.3}{V} \right)}
⇒[H(aq)+]=(2×10−9)×(0.50.3)=3.33×10−9mol/l\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 pHpH value in equation;
pH=−log⁡[H(aq)+]pH=-\log [H_{(aq)}^{+}]
∴pH=−log⁡[3.33×10−9]=8.47\therefore pH=-\log [3.33\times {{10}^{-9}}]=8.47

Note :
The pH scale generally ranges from 0 to 140\text{ }to\text{ }14 . It helps us to determine how acidic or basic the solution is. If the pHpH of the solution is less than 77 , then it is acidic and if the pHpH of the solution is more than 77 , then it is the basic solution. Similarly, if the pHpH of the solution of equal to 77 , then the solution is neutral.