Question

To maintain the pH of \[7.4\] for blood at normal condition which is \[2M\] in .what volume of \[5M\] of \[NaHC{O_3}\] solution is required to be mixed with\[10\] ml of blood?
Pk(a1) \[{H_2}C{O_3}\]=\[6.4\] at a given temperature.

Answer 1

Hint:
In common language pH is called power of hydrogen. It is basically nothing but the scale which is used to specify the acidity and basicity of an aqueous solution. \[5M\] of \[NaHC{O_3}\] solution
the pH of \[7.4\] for blood
at normal condition which is \[2M\]
we need to find the volume to mix with\[10\] ml of blood
then writing the equation of finding Ph :
\[pH = - \log {K_a} + \log \dfrac{{NaHC{O_3}}}{{{H_2}C{O_3}}}\]

Complete step-by-step answer: In common language pH is called power of hydrogen. It is basically nothing but the scale which is used to specify the acidity and basicity of an aqueous solution. When we further talk about the acidity or basicity then it is very simple to remember that acidic solutions are none other than the solutions which contain higher concentration of hydrogen ions. Further talking about the basicity or the basic solutions are the solutions which have higher concentration of the hydroxide ions. When we consider the technical terms then it is nothing but the logarithmic and inversely indicates the concentration of hydrogen which we have already discussed.
We also further know that the solutions are acidic if they have their Ph less than seven and the solutions are basic in nature if they have their Ph greater than seven. A neutral solution has a Ph of seven itself.
Talking about the question we have
\[5M\] of \[NaHC{O_3}\] solution
the pH of \[7.4\] for blood
at normal condition which is \[2M\]
we need to find the volume to mix with\[10\]ml of blood
then writing the equation of finding Ph :
\[pH = - \log {K_a} + \log \dfrac{{NaHC{O_3}}}{{{H_2}C{O_3}}}\]
Then we should calculate the concentration of both :
\[NaHC{O_3}\]=\[\dfrac{{5 \times V}}{{(V + 10)}}M\]
\[{H_2}C{O_3}\]= \[\dfrac{{2 \times 10}}{{V + 10}}M\]
Where \[V + 10\] is the total volume and V is to be calculated as by putting the values:
\[7.4 = - \log 7.8 \times {10^{ - 7}} + \log \dfrac{{5 \times V}}{{V + 10}} \times {\dfrac{{V + 10}}{
2 \times 10 }}\]
\[V = 78.32ml\]

Note: ph should be calculated by suitably putting the valued into the equation. Further each quantity should be in the standard form.