Calculating pH and Ionization of Aspirin Solutions

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To calculate the pH of a 0.2M aspirin solution, the formula pH = -log[H+] is used, with [H+] derived from the equation [H+] = √(Ka * c), resulting in a pH of approximately 3.52. The percent ionization can be determined using the formula √(Ka/c), which quantifies how much aspirin dissociates in solution. Adding 0.01M hydrochloric acid increases the concentration of H+, affecting the ionization of aspirin due to the common ion effect, which suppresses further dissociation of the weak acid. The new pH and percent ionization must be recalculated using the updated H+ concentration from the combined effects of aspirin and hydrochloric acid. This discussion emphasizes the calculations and qualitative effects of ionization in weak acid solutions.
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Question: Aspirin is a weak acid.

(a) Calculate pH of 0.2M solution of aspirin at 25 degrees celsius (Ka = 3.0 x 10^-4 at 25 degrees celcius).

(b) Determine the percent ionisation

(c) Explain qualitatively the effect of adding 0.01M hydrochloric acid to the aspirin solution.

(d) Calculate the pH of the resulting solution

(e) Determine the percent ionisation of the aspirin under these conditions
 
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hmmm
well formula goes.. pH=-log[H+]
am not sure what about aspirine...i am is Ka also meant as concentration of H+ ??
if yes ...then

pH=-log[H+]
pH=-log(3.0 x 10^-4)=3.5228

and it is acid...but wouldn't say that weak
 
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here is the most effective way with which u must proceed---

1) the conc is given as .2 M and ka is given so find the H^+ conc using the formula [H^+] = \sqrt{K_a * c}
2) percentage ionisation is given by the formula \sqrt{\frac{k_a}{c}}
3)Explain common ion effect
4) due to addition of hydrochloric acid the conc of H+ will now be 0.01 + the conc of H+ u got in the 1st question
5) repeate the 2 question but now with the new conc of H+ note use formula \frac{[H^+]}{c}
 
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