Question

A solution is made by mixing 50 ml of 2.0m k2hpo4 and 25 ml of 2.0m kh2po4. the solution is diluted to a final volume of 200 ml. what is the ph of the final solution

Answer 1

50 mL of 2.0 M of $K_2HPO_4$ and 25 mL of 2.0 M of $KH_2PO_4$ were mixed to make a solution

Final volume of the solution after dilution = 200 mL    (given)

Final concentration of $K_2HPO_4$, [$K_2HPO_4$] = $\frac{50 mL\times 2 M}{200 mL} = 0.5 M$

Final concentration of $KH_2PO_4$, [$KH_2PO_4$] = $\frac{25 mL\times 2 M}{200 mL} = 0.25 M$

Using Hasselbach- Henderson equation:

$pH = pK_a+ log \frac{[salt]}{[acid]}$

$pka of KH_2PO_4 = 6.85$

Substituting the values:

$pH = 6.85+ log \frac{0.5}{0.25}$

$pH = 6.85+ log 2$

$pH = 6.85+ 0.3 = 7.15$

Hence,  the $pH$ of the final solution is 7.15.

Answer 2

The pH of the final solution is $\boxed{7.15}$ .

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:

${{\text{M}}_{{\text{conc}}}}{{\text{V}}_{{\text{conc}}}} = {{\text{M}}_{{\text{dil}}}}{{\text{V}}_{{\text{dil}}}}$        ...... (1)                                                              

Here,

${{\text{M}}_{{\text{conc}}}}$ is the molarity of the concentrated solution.

${{\text{V}}_{{\text{conc}}}}$ is the volume of the concentrated solution.

${{\text{M}}_{{\text{dil}}}}$ is the molarity of a dilute solution.

${{\text{V}}_{_{{\text{dil}}}}}$ is the volume of dilute solution.

Rearrange equation (3) to calculate ${{\text{M}}_{_{{\text{dil}}}}}$.

${{\text{M}}_{{\text{dil}}}} = \dfrac{{{{\text{M}}_{{\text{conc}}}}{{\text{V}}_{{\text{conc}}}}}}{{{{\text{V}}_{{\text{dil}}}}}}$   …… (2)                                                                      

Substitute 2 M for ${{\text{M}}_{{\text{conc}}}}$, 50 mL for ${{\text{V}}_{{\text{conc}}}}$, 200 mL for ${{\text{V}}_{_{{\text{dil}}}}}$ in equation (2) to calculate the concentration of ${{\text{K}}_2}{\text{HP}}{{\text{O}}_4}$.

$\begin{aligned}{{\text{M}}_{{{\text{K}}_2}{\text{HP}}{{\text{O}}_4}}} &= \frac{{\left( {2{\text{ M}}} \right)\left( {50{\text{ mL}}} \right)}}{{{\text{200 mL}}}}\\&= 0.5{\text{ M}}\\\end{aligned}$  

Substitute 2 M for ${{\text{M}}_{{\text{conc}}}}$, 25 mL for ${{\text{V}}_{{\text{conc}}}}$, 200 mL for ${{\text{V}}_{_{{\text{dil}}}}}$ in equation (2) to calculate the concentration of ${\text{K}}{{\text{H}}_2}{\text{P}}{{\text{O}}_4}$.

 $\begin{aligned}{{\text{M}}_{{\text{K}}{{\text{H}}_2}{\text{P}}{{\text{O}}_4}}} &= \frac{{\left( {2{\text{ M}}} \right)\left( {25{\text{ mL}}} \right)}}{{{\text{200 mL}}}}\\&= 0.25{\text{ M}}\\\end{aligned}$

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:

${\text{pH}} = {\text{p}}{K_{\text{A}}} + {\text{log}}\dfrac{{\left[ {{\text{Salt}}} \right]}}{{\left[ {{\text{Acid}}} \right]}}$  

In the given system, ${{\text{K}}_2}{\text{HP}}{{\text{O}}_4}$ is a salt whereas ${\text{K}}{{\text{H}}_2}{\text{P}}{{\text{O}}_4}$ is an acid. So Henderson-Hasselbalch equation for the given system is as follows:

${\text{pH}} = {\text{p}}{K_{\text{A}}} + {\text{log}}\dfrac{{\left[ {{{\text{K}}_2}{\text{HP}}{{\text{O}}_4}} \right]}}{{\left[ {{\text{K}}{{\text{H}}_2}{\text{P}}{{\text{O}}_4}} \right]}}$           …… (3)                                                

The value of ${\text{p}}{K_{\text{A}}}$ is 6.85.

The value of $\left[ {{{\text{K}}_2}{\text{HP}}{{\text{O}}_4}} \right]$ is 0.5 M.

The value of $\left[ {{\text{K}}{{\text{H}}_2}{\text{P}}{{\text{O}}_4}} \right]$ is 0.25 M.

Substitute these values in equation (3).

$\begin{aligned}{\text{pH}} &= 6.85 + {\text{log}}\left( {\frac{{0.5{\text{ M}}}}{{0.25{\text{ M}}}}} \right) \\&= 7.15\\\end{aligned}$  

Learn more:

1. The reason for the acidity of water brainly.com/question/1550328
2. Reason for the acidic and basic nature of amino acid. 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.