Respuesta :
50 mL of 2.0 M of [tex]K_2HPO_4[/tex] and 25 mL of 2.0 M of [tex]KH_2PO_4[/tex] were mixed to make a solution
Final volume of the solution after dilution = 200 mL (given)
Final concentration of [tex]K_2HPO_4[/tex], [[tex]K_2HPO_4[/tex]] = [tex]\frac{50 mL\times 2 M}{200 mL} = 0.5 M[/tex]
Final concentration of [tex]KH_2PO_4[/tex], [[tex]KH_2PO_4[/tex]] = [tex]\frac{25 mL\times 2 M}{200 mL} = 0.25 M[/tex]
Using Hasselbach- Henderson equation:
[tex]pH = pK_a+ log \frac{[salt]}{[acid]}[/tex]
[tex]pka of KH_2PO_4 = 6.85[/tex]
Substituting the values:
[tex]pH = 6.85+ log \frac{0.5}{0.25}[/tex]
[tex]pH = 6.85+ log 2[/tex]
[tex]pH = 6.85+ 0.3 = 7.15[/tex]
Hence, the [tex]pH[/tex] of the final solution is 7.15.
The pH of the final solution is [tex]\boxed{7.15}[/tex] .
Further Explanation:
When extra amount of solvent is added, it results into dilution of concentrated solution but the amount of solute remains the same. There is a change in the volume of solution once dilution is done.
The following expression relates molarity and volume of dilute and concentrated solutions:
[tex]{{\text{M}}_{{\text{conc}}}}{{\text{V}}_{{\text{conc}}}} = {{\text{M}}_{{\text{dil}}}}{{\text{V}}_{{\text{dil}}}}[/tex] ...... (1)
Here,
[tex]{{\text{M}}_{{\text{conc}}}}[/tex] is the molarity of the concentrated solution.
[tex]{{\text{V}}_{{\text{conc}}}}[/tex] is the volume of the concentrated solution.
[tex]{{\text{M}}_{{\text{dil}}}}[/tex] is the molarity of a dilute solution.
[tex]{{\text{V}}_{_{{\text{dil}}}}}[/tex] is the volume of dilute solution.
Rearrange equation (3) to calculate [tex]{{\text{M}}_{_{{\text{dil}}}}}[/tex].
[tex]{{\text{M}}_{{\text{dil}}}} = \dfrac{{{{\text{M}}_{{\text{conc}}}}{{\text{V}}_{{\text{conc}}}}}}{{{{\text{V}}_{{\text{dil}}}}}}[/tex] …… (2)
Substitute 2 M for [tex]{{\text{M}}_{{\text{conc}}}}[/tex], 50 mL for [tex]{{\text{V}}_{{\text{conc}}}}[/tex], 200 mL for [tex]{{\text{V}}_{_{{\text{dil}}}}}[/tex] in equation (2) to calculate the concentration of [tex]{{\text{K}}_2}{\text{HP}}{{\text{O}}_4}[/tex].
[tex]\begin{aligned}{{\text{M}}_{{{\text{K}}_2}{\text{HP}}{{\text{O}}_4}}} &= \frac{{\left( {2{\text{ M}}} \right)\left( {50{\text{ mL}}} \right)}}{{{\text{200 mL}}}}\\&= 0.5{\text{ M}}\\\end{aligned}[/tex]
Substitute 2 M for [tex]{{\text{M}}_{{\text{conc}}}}[/tex], 25 mL for [tex]{{\text{V}}_{{\text{conc}}}}[/tex], 200 mL for [tex]{{\text{V}}_{_{{\text{dil}}}}}[/tex] in equation (2) to calculate the concentration of [tex]{\text{K}}{{\text{H}}_2}{\text{P}}{{\text{O}}_4}[/tex].
[tex]\begin{aligned}{{\text{M}}_{{\text{K}}{{\text{H}}_2}{\text{P}}{{\text{O}}_4}}} &= \frac{{\left( {2{\text{ M}}} \right)\left( {25{\text{ mL}}} \right)}}{{{\text{200 mL}}}}\\&= 0.25{\text{ M}}\\\end{aligned}[/tex]
Henderson-Hasselbalch equation is a mathematical expression that is used to determine the of buffer solutions. The general expression for pH of buffer solution is as follows:
[tex]{\text{pH}} = {\text{p}}{K_{\text{A}}} + {\text{log}}\dfrac{{\left[ {{\text{Salt}}} \right]}}{{\left[ {{\text{Acid}}} \right]}}[/tex]
In the given system, [tex]{{\text{K}}_2}{\text{HP}}{{\text{O}}_4}[/tex] is a salt whereas [tex]{\text{K}}{{\text{H}}_2}{\text{P}}{{\text{O}}_4}[/tex] is an acid. So Henderson-Hasselbalch equation for the given system is as follows:
[tex]{\text{pH}} = {\text{p}}{K_{\text{A}}} + {\text{log}}\dfrac{{\left[ {{{\text{K}}_2}{\text{HP}}{{\text{O}}_4}} \right]}}{{\left[ {{\text{K}}{{\text{H}}_2}{\text{P}}{{\text{O}}_4}} \right]}}[/tex] …… (3)
The value of [tex]{\text{p}}{K_{\text{A}}}[/tex] is 6.85.
The value of [tex]\left[ {{{\text{K}}_2}{\text{HP}}{{\text{O}}_4}} \right][/tex] is 0.5 M.
The value of [tex]\left[ {{\text{K}}{{\text{H}}_2}{\text{P}}{{\text{O}}_4}} \right][/tex] is 0.25 M.
Substitute these values in equation (3).
[tex]\begin{aligned}{\text{pH}} &= 6.85 + {\text{log}}\left( {\frac{{0.5{\text{ M}}}}{{0.25{\text{ M}}}}} \right) \\&= 7.15\\\end{aligned}[/tex]
Learn more:
- The reason for the acidity of water https://brainly.com/question/1550328
- Reason for the acidic and basic nature of amino acid. https://brainly.com/question/5050077
Answer details:
Grade: High School
Subject: Chemistry
Chapter: Acid, base and salts.
Keywords: dilution, concentrated solution, pH, pKA, 0.5 M, 0.25 M, K2HPO4, KH2PO4, 6.85, 200 mL, 50 mL, 25 mL.
