Calculate the pH of a buffer solution composed of 0.070 mol ·L^-1 trichloroacetic acid (CCl3COOH

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Calculate the pH of a buffer solution composed of 0.070 mol...

Calculate the pH of a buffer solution composed of 0.070 mol \cdot L^{-1} trichloroacetic acid ($CCl_3COOH$) and 0.015 mol \cdot L^{-1} sodium trichloroacetate ($CCl_3COONa$) by using the equilibrium method. The $K_a$ for $CCl_3COOH$ is $2.2 \times 10^{-1}$. Express your answer to two decimal places. $pH = $ Part B Calculate the pH of a buffer solution composed of 0.070 mol$\cdot L^{-1}$ trichloroacetic acid ($CCl_3COOH$) and 0.015 mol$\cdot L^{-1}$ sodium trichloroacetate ($CCl_3COONa$) by using the Henderson-Hasselbalch equation. The $K_a$ for $CCl_3COOH$ is $2.2 \times 10^{-1}$. Express your answer to two decimal places. $pH = $

Transcript

00:01 Initial volume that is v1 can be written as 55 .2 ml which is given initial concentration is also given that is 0 .221 molar this is your initial condition so final volume final volume denoted by here i will denoted with v2 is equals to will 55 .2 plus 13 .2 which will be equals to 68 .4 ml this will be the final volume and we have to find out the final concentration c2 so by using the law of equivalence we will write c1 v1 concentration is actually your molarity okay so initial volume will be equals to do initial uh concentration into initial volume will be equals to final concentration into final volume so from here initial concentration so final concentration will be equals to initial concentration into initial volume divided by final volume now put all the values and calculate for the value of final concentration which will be equals to 55 .2 which is initial volume into initial concentration 0 .221 divided by final volume 68 .4 so your final concentration is equal to 0 .178 molar this is your final concentration now concentration of h c h c h .o2 is equals to 0 .178 molar this is the final concentration of your methenoic asset.

01:36 Similarly for naoh, n .a .o .2 ch, it is not naoh, it is nacchocho.

01:54 The concentration of nacho2 in mixture will be equals to 13 .2 into 0 .60, divided by that is number of moles divided by total volume that is 68 .4 which will be equals to 0 .108 molar okay divided it 1000 because we have to take volume in liter to calculate concentration so mixture of mixture of h c h o2 a weak acid and and and h -c -o -2 minus the conjugate base, which is the conjugate base, forms a buffer solution.

03:05 Forms a buffer solution.

03:12 So h -c -h -o -2 forms buffer solution with h -plus plus c -h -o -2 minus...

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