Question

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

Answer 1

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 $HNO_2$ = 0.55 M.

• Concentration of $KNO_2$ = 0.75 M.

• Rate constant = $6.8 \times 10^{-4}$

How to determine the pH of the buffer solution.

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

                                     $HNO_2(aq)\rightleftharpoons H^{2+} (aq)+ KNO_2^{-}(aq)$

Initial cond.                       0.55M              0          0.75M

    $\Delta C$                                   -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:

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

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

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

pH = 3.30.

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

$pH =pka+ log_{10} \frac{A^-}{HA}$

Where:

• HA is acetic acid.

• $A^-$  is acetate ion.

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