Answered

Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly 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 your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.