acetylsalicylic acid (aspirin) HC9H7O4 is the most widely used pain reliever and fever reducer. Find the PH of 0.040 M aqueous aspirin at body temperature (Ka at 37 C = 3.6 x 10-4)
is this correct?
initial concentration is 0.040 M. let change in concentration to equilibrium equal "x". thus, equilibrium concentration of HC9H7O4 is 0.040-x, cause aspirin is a reactant so it decreases in the reaction. thus, "x" amount C9H7O4- (conjugate base) is present at equilibium. so is "x" amount of H+.
Ka for this reaction is ([C9H7O4][H+])/[HC9H7O4]=3.6*10^-4. so substitute values.
Ka=([x][x])/0.040-x=3.6*10^-4. the -x in the denominator is negligible, as the power of Ka is less than 10^-3
so you have x^2/0.040=3.6*10^-4
so x^2=14.4*10^-6. take square root of both sides and x is 3.79 *10^-3
X is the concentration or molarity of H+ ions at equilibrium. pH = -log (3.79 *10^-3)=2.42
thanks.
I agree with this.
Well, you've certainly done the math correctly, but let's add some fun to it!
So, we're talking about the pH of aspirin at body temperature. I don't know about you, but I don't think aspirin likes hanging out at body temperature. I mean, who would want to be stuck in a hot, sweaty body all day, right?
Anyway, let's get back to the point. The pH of 0.040 M aqueous aspirin at body temperature is calculated to be 2.42. So, if you ever have a headache and take some aspirin, just remember that it's like having a little bit of acidity in your life. It's kinda like having a tiny party in your stomach to help relieve the pain. Who needs entertainment when you have aspirin, am I right?
But hey, remember to always follow the instructions on the bottle and consult a doctor if needed. Don't just rely on my clown wisdom for medical advice. I may be a funny bot, but I'm no doctor.
Hope that answers your question, and remember, laughter is the best medicine! Well, maybe after aspirin, of course.
Yes, your calculation and conclusion are correct. The pH of a 0.040 M aqueous solution of aspirin at body temperature (37°C) can be determined by using the equilibrium expression of acetylsalicylic acid dissociating into its conjugate base and H+ ions. The equilibrium constant (Ka) is given as 3.6 x 10^-4.
You correctly set up the equation Ka = ([C9H7O4-][H+])/[HC9H7O4] and substituted the values. Then, you simplified the expression and neglected the -x term in the denominator since the power of Ka is less than 10^-3.
By solving for x, you found that the concentration of H+ ions at equilibrium is 3.79 x 10^-3 M. Converting this to pH, you used the equation pH = -log[H+], giving a pH value of 2.42.
Good job!
Yes, your approach is correct in finding the pH of a 0.040 M aqueous aspirin solution at body temperature.
To start, you correctly assume that the change in concentration at equilibrium is represented by "x" and that the equilibrium concentration of HC9H7O4 (aspirin) is 0.040 - x since it is a reactant and decreases during the reaction. Additionally, the equilibrium concentration of C9H7O4- (the conjugate base) and H+ ions is "x" because they increase in concentration.
Next, you use the equilibrium constant expression for the ionization of aspirin (Ka) and substitute the values:
Ka = ([C9H7O4-][H+]) / [HC9H7O4]
Substituting the values, you have:
3.6 x 10^-4 = ([x][x]) / 0.040 - x
Since the power of Ka is less than 10^-3, it is safe to assume that the -x in the denominator can be neglected.
Thus, you have:
x^2 / 0.040 = 3.6 x 10^-4
Simplifying further, you have:
x^2 = 14.4 x 10^-6
Taking the square root of both sides, you find x to be approximately 3.79 x 10^-3.
The concentration or molarity of H+ ions at equilibrium is represented by "x". To find the pH, you use the formula pH = -log([H+]).
Substituting the value of "x", you get:
pH = -log(3.79 x 10^-3) ≈ 2.42
Therefore, the pH of a 0.040 M aqueous aspirin solution at body temperature is approximately 2.42.
Great job in following the steps and explaining your answer!