
What is the pH of $ 3.1 \times {10^{ - 3}}M\,\,\,\,HCl? $
Answer
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Hint: In this question we have to find the pH of the given data. We know that the degree of alkalinity or acidity of a solution is known as its pH . “pH” is the negative algorithm of hydrogen ion concentration. Here we have been given the concentration of $ HCl $ solution. From the concentration of $ HCl $ we will calculate the concentration of hydrogen ions, then we will calculate the pH.
Complete answer:
We know that if the pH of the solution is less than $ 7 $ , then the solution has an acidic nature. If the pH of the solution is more than $ 7 $ then the solution has basic nature. If the pH of the solution is equal to $ 7 $ , then we know that the solution is neither acidic or basic, it is neutral in nature .
Here we have been given the concentration of $ HCl $ which is $ 3.1 \times {10^{ - 3}}M $ .
We know that $ HCl $ is a strong acid and dissociates completely in water.
The dissociation reaction of $ HCl $ is as follow:
$ HCl \xrightarrow H^+ + Cl^- $
We know that the concentration of $ {H^ + } $ is equal to $ HCl $
So we can say that the concentration of $ {H^ + } $ due to $ HCl $ is
$ 3.1 \times {10^{ - 3}}M $ .
We know that pH is the negative logarithm of the hydrogen ion concentration, so we can write this as $ pH = - \log \left[ {{H^ + }} \right] $
We will substitute $ 3.1 \times {10^{ - 3}}M $ for the hydrogen ion concentration, Thus we have:
$ pH = - \log \left( {\;3.1 \times {{10}^{ - 3}}M} \right) $
We will now simplify this:
$ \Rightarrow pH = - \log \left( {\;\dfrac{{31}}{{10}} \times \dfrac{1}{{1000}}M} \right) $
$ \Rightarrow pH = - \log \left( {\;0.0031} \right) $
Now we will substitute the value of given logarithm expression and we have:
$ \Rightarrow pH = - ( - 2.5086) $
$ \Rightarrow pH = + 2.5086 $
Hence the required pH is $ 2.51 $ .
Note:
We should note that as we have calculated above the pH of the hydrogen ion is less than $ 7 $ , so we can say that the $ 3.1 \times {10^{ - 3}}M $ solution of $ HCl $ is acidic in nature. We must always remember that as concentration increases with temperature, the pH always changes. With the increase in temperature the pH decreases, however this does not mean that the water becomes acidic at higher temperature.
Complete answer:
We know that if the pH of the solution is less than $ 7 $ , then the solution has an acidic nature. If the pH of the solution is more than $ 7 $ then the solution has basic nature. If the pH of the solution is equal to $ 7 $ , then we know that the solution is neither acidic or basic, it is neutral in nature .
Here we have been given the concentration of $ HCl $ which is $ 3.1 \times {10^{ - 3}}M $ .
We know that $ HCl $ is a strong acid and dissociates completely in water.
The dissociation reaction of $ HCl $ is as follow:
$ HCl \xrightarrow H^+ + Cl^- $
We know that the concentration of $ {H^ + } $ is equal to $ HCl $
So we can say that the concentration of $ {H^ + } $ due to $ HCl $ is
$ 3.1 \times {10^{ - 3}}M $ .
We know that pH is the negative logarithm of the hydrogen ion concentration, so we can write this as $ pH = - \log \left[ {{H^ + }} \right] $
We will substitute $ 3.1 \times {10^{ - 3}}M $ for the hydrogen ion concentration, Thus we have:
$ pH = - \log \left( {\;3.1 \times {{10}^{ - 3}}M} \right) $
We will now simplify this:
$ \Rightarrow pH = - \log \left( {\;\dfrac{{31}}{{10}} \times \dfrac{1}{{1000}}M} \right) $
$ \Rightarrow pH = - \log \left( {\;0.0031} \right) $
Now we will substitute the value of given logarithm expression and we have:
$ \Rightarrow pH = - ( - 2.5086) $
$ \Rightarrow pH = + 2.5086 $
Hence the required pH is $ 2.51 $ .
Note:
We should note that as we have calculated above the pH of the hydrogen ion is less than $ 7 $ , so we can say that the $ 3.1 \times {10^{ - 3}}M $ solution of $ HCl $ is acidic in nature. We must always remember that as concentration increases with temperature, the pH always changes. With the increase in temperature the pH decreases, however this does not mean that the water becomes acidic at higher temperature.
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