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Saccharin, $\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{NSO}_{3} \mathrm{H},$ is a weak acid $\left(K_{\mathrm{a}}=2.1 \times 10^{-2}\right) .$ If 0.250 $\mathrm{L}$ of diet cola with a buffered pH of 5.48 was prepared from $2.00 \times 10^{-3} \mathrm{g}$ of sodium sacharide, $\mathrm{Na}\left(\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{NSO}_{3}\right),$ what are the final concentrations of saccharine and sodium saccharide in the solution?
$2.5 \times 10^{-11} \mathrm{M}$
Chemistry 102
Chapter 14
Acid-Base Equilibria
Liquids
University of Maryland - University College
University of Toronto
Lectures
03:07
A liquid is a nearly incom…
04:38
A liquid is a state of mat…
04:36
Saccharin (HC $_{7} \mathr…
03:48
Saccharin $\left(\mathrm{H…
05:20
Saccharin, a sugar substit…
02:46
04:11
You have $100.0 \mathrm{g}…
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In the case of very weak a…
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Saccharin $\left(\mathrm{C…
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The $K_{\mathrm{a}}$ of th…
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Quinolinic acid, $\mathrm{…
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Let's determine the concentration. Final concentration of saccharine and sodium saturated This buffer solution. We're told that a k a. For the saccharine weak acid C seven each four. 10 s three h is equal to 2.1 times 10 to the negative, too. Calculate the P k A of this on it's equal the negative log 2.1. I'm saying the negative, too, and we find that that is equal to 1.67 were given the pH of the buffer so we can use the Henderson household back equation. The P H is equal to the peak a plus log of the concentration of the base over the concentration of country acid. And we can substitute this further. Ph is equal to the peak a plus the log of the base, which is the A C seven h four and s 03 over the acid, which is si seven h four n s, 03 h. Substitutes and values in here find that 5.48 is the pH. The P k K was 1.67 plus even log of N a C seven h four. An asshole. Three. Hopefully, concentration of the acid and rearranging. We get a relationship Where the log of the n a C seven h four and s 03 over D C seven h four n s 03 h would be equal to 3.81 And applying her, um, log function here the morality of the A si seven h four n s, 03 over C seven h four n s 03 h the equal Attend to the S 31 exploded 381 which is equal to, uh, 6457. We'll come back to this in a minute. Let's now find the moles of the and A C seven h four n s 03 We're told that we're starting with 20 times 10 to the negative three grounds. Moeller Mass is 205.17 g per mole and this will work out to 9.75 times 10 to the negative six moles. And now we know the volumes. We could find the polarity of the A C seven h four and s 03 9.75 times 10 to the negative six moles. Morality is in the question 0.250 Leaders of the Diet Cola 0.250 leaders We find that this is equal to three 0.89 times 10 to the negative five. Mueller. So here's the morality of the sodium, um, sack ride and going back to the equation from up above here we want the malaria t of D. C. Seven h four s, 03 and that's equal to um, right the mill arat e of the rearranging and a C seven h four n s 0 3/6 outs of 457 3.89 times 10 to the negative. Five. Mueller divided by 6457. We find that the morality of the asset is 6.2 times 10 to the negative nine. Mueller. So here's our morality of the saccharine and up above. We have the morality of the sodium sacrifice
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