Question

: 500 ml of \[{\text{0}}{\text{.2 M BOH}}\] (a weak base) is mixed with 500 ml of \[{\text{0}}{\text{.1 M HCl}}\] and pH of resulting solution is 9. What is the pH of a \[{\text{0}}{\text{.1 M BCl}}\] solution? \[\left[ {{\text{antilog}}\left( { - 9} \right) = {{10}^{ - 9}}} \right]\]

Answer 1

: Hint:We need to do the neutralization of an acid and base. Here due to the presence of a weak base and its salt a basic buffer will form. The value of equilibrium constant for base can be calculated using the formula of buffer. We can calculate the pH of salt by salt hydrolysis.
Formula used:
\[{\text{pOH}} = {\text{p}}{{\text{K}}_{\text{b}}} + \log \dfrac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{base}}} \right]}}\]
Here \[{{\text{K}}_{\text{b}}}\] is the equilibrium constant for a base.
For pH calculation of weak base and strong acid: \[{\text{pH}} = \dfrac{1}{2}\left[ {{\text{p}}{{\text{K}}_{\text{w}}} - {\text{p}}{{\text{K}}_{\text{b}}} - \log {\text{c}}} \right]\]
Here \[{{\text{K}}_{\text{w}}}\] is the ionic product of water and C is the concentration of salt.

Complete step by step answer:
When an acid and base is present the neutralization reaction occurs and a salt if formed. Let us first calculate the millimoles of acid and base present:
Millimoles of \[{\text{0}}{\text{.2 M BOH}}\] is \[0.2 \times 500 = 100\]
Millimoles of \[{\text{0}}{\text{.1 M HCl}}\] is \[{\text{0}}{\text{.1 }} \times {\text{500}} = 50\]
Hence 50 milli mole of acid and 50 millimoles of base will react to form 50 millimoles of salt and 50 milli mole of base will remain in solution. The total volume becomes 1000 mL.
Concentration of base will be \[\dfrac{{50}}{{1000}} = 0.05{\text{ M}}\]
Concentration of salt will be \[\dfrac{{50}}{{1000}} = 0.05{\text{ M}}\]
pH is given to us that is 9 so pOH will be \[14 - 9 = 5\], using the formula we will get:
\[{\text{5}} = {\text{p}}{{\text{K}}_{\text{b}}} + \log \dfrac{{{\text{0}}{\text{.05}}}}{{0.05}}\]
Log 1 is zero and p is the operator for – log.
\[ - \log {{\text{K}}_{\text{B}}} = 5\]
\[{{\text{K}}_{\text{B}}} = {10^{ - 5}}\].
Now we need to calculate the pH for a \[{\text{0}}{\text{.1 M BCl}}\]. Using the formula we will get:
\[{\text{pH}} = \dfrac{1}{2}\left[ { - \log {{10}^{ - 14}} + \log {{10}^{ - 5}} - \log 0.1} \right]\]
\[ \Rightarrow {\text{pH}} = \dfrac{1}{2}\left[ {14 - 5 + 1} \right] = 5\]

Note:
pH is a scale that is used to define the concentration of hydrogen ions. Since the concentration of hydrogen in terms of molarity is very less so we devised a new method known as pH which gives us the value in numbers and is easy to remember and work on. The pH is known as the power of hydrogen. p is a mathematical operator that is –log.