Answered

Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

1) A solution is prepared with 0.55 M HNO2 and 0.75 M KNO2. Fill in the ICE Table with the appropriate values

2) The Ka for this solution of HNO2 is 6.8x10^-4. Based on the ICE table, set up the expression for Ka.

3) Based on your ICE table, determine the pH of the buffer solution

1 A Solution Is Prepared With 055 M HNO2 And 075 M KNO2 Fill In The ICE Table With The Appropriate Values 2 The Ka For This Solution Of HNO2 Is 68x104 Based On class=
1 A Solution Is Prepared With 055 M HNO2 And 075 M KNO2 Fill In The ICE Table With The Appropriate Values 2 The Ka For This Solution Of HNO2 Is 68x104 Based On class=
1 A Solution Is Prepared With 055 M HNO2 And 075 M KNO2 Fill In The ICE Table With The Appropriate Values 2 The Ka For This Solution Of HNO2 Is 68x104 Based On class=

Sagot :

Lanuel

Based on the calculations through the ICE table, the pH of the buffer solution is equal to 3.30.

Given the following data:

  • Concentration of [tex]HNO_2[/tex] = 0.55 M.
  • Concentration of [tex]KNO_2[/tex] = 0.75 M.
  • Rate constant = [tex]6.8 \times 10^{-4}[/tex]

How to determine the pH of the buffer solution.

First of all, we would write the properly balanced chemical equation for this chemical reaction:

                                     [tex]HNO_2(aq)\rightleftharpoons H^{2+} (aq)+ KNO_2^{-}(aq)[/tex]

Initial cond.                       0.55M              0          0.75M

    [tex]\Delta C[/tex]                                   -x                   x               x

At equib.                           0.55M - x         0 + x      0.75M + x

From the ICE table, the Ka for this chemical reaction is given by:

[tex]K_{a}=\frac{[H^+][NO_2^+]}{HNO_2} \\\\H^+ = K_{a}\frac{[HNO_2]}{[NO_2^+]} \\\\H^+ = 6.8 \times 10^{-4} \times \frac{0.55}{0.75} \\\\H^+ = 6.8 \times 10^{-4} \times 0.733\\\\H^+ = 4.98 \times 10^{-4} \;M[/tex]

Now, we can calculate the pH of the buffer solution:

[tex]pH=-log[H^+]\\\\pH=-log[4.98 \times 10^{-4}]\\\\pH=-(-3.30)[/tex]

pH = 3.30.

Alternatively, you can calculate the pH of this buffer solution by applying Henderson-Hasselbalch equation:

[tex]pH =pka+ log_{10} \frac{A^-}{HA}[/tex]

Where:

  • HA is acetic acid.
  • [tex]A^-[/tex]  is acetate ion.

Read more on concentration here: https://brainly.com/question/3006391

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.