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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

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