I need to find out what the total acid concentration is in Kool-Aid.
If 23.3 mL of 0.1043M NaOH neutralize a 60 mL solution of Kool-Aid (10 mL) and water (50 mL), what is the total concentration in a 10.00 mL aliquot of Kool-Aid? If you prepared a standard 250 mL serving by mixing a packet of Kool-Aid with 2 L of water, what would the total acid concentration be?
What I've probably calculated incorrectly so far...
23.3 mL NaOH x 0.1043 M NaOH = 2.430 mMol NaOH. = 1 mole OH = 2.430 mMol H+.
2.430 mMol H+/60 mL = 0.040 M H+.
For the total acid concentration in a 10 mL sample,
(0.040 x 0.06)/0.01 = 0.24 M
For the total acid concentration in a 250 mL serving,
(0.040 x 0.06)/2 = 1.2 x 10^-3
The balanced equation is:
5 citric acid + 2 ascorbic acid + 17 NaOH -> 5 citrate ion + 2 ascorbic acid + 17 H20.
The equation of interest is: H+ + OH- -> H2O
Any help would be appreciated.
Thanks
I can help with the first part. The second part is unclear exactly how the sample was prepared and there is no data given to determine the amount of sample in the packet.
The first part is
mL x M = mmoles and you are correct in
23.3 mL x 0.1043M = 2.430 mmoles. However, you don't divide by 60. The entire 10 mL portion was titrated; therefore, 2.430 mmoles is the amount in the 10 mL. It is unclear what this 10.00 mL represents; i.e., is it a solution prepared (and how) or is it a liquid concentrate of kool-aid?
To find the total acid concentration in Kool-Aid, you need to determine the concentration of H+ ions in the solution. In this case, you can use the neutralization reaction between NaOH and H+ to find the concentration.
You correctly calculated the number of moles of NaOH (2.430 mMol) and the number of moles of H+ ions (also 2.430 mMol) based on the balanced equation. However, when determining the concentration, you made an error in your calculation.
To find the concentration (Molarity) of H+ ions, you divide the number of moles of H+ by the volume in liters. In this case, the volume is 60 mL or 0.06 L. So the correct calculation is:
2.430 mMol / 0.06 L = 40 mMol/L = 0.040 M
Therefore, the total concentration of H+ ions in the 10.00 mL aliquot of Kool-Aid is 0.040 M.
Now, let's calculate the total acid concentration in a 250 mL serving of Kool-Aid.
Since the 250 mL serving is prepared by mixing a packet of Kool-Aid with 2 L (2000 mL) of water, you need to determine the concentration based on the dilution.
The number of moles of H+ ions remains the same (2.430 mMol), but the volume increases to 2000 mL + 10 mL (from the packet) = 2010 mL or 2.01 L.
Now, calculate the concentration:
2.430 mMol / 2.01 L = 1.2 mMol/L = 1.2 × 10^-3 M
Therefore, the total acid concentration in a 250 mL serving of Kool-Aid is 1.2 × 10^-3 M.
Remember to double-check your calculations and the assumptions made about the composition of Kool-Aid.
Your calculations seem correct. Here is the step-by-step breakdown of the solution:
1. Calculate the number of moles of NaOH used:
23.3 mL NaOH x 0.1043 M NaOH = 2.430 mmol NaOH
2. Since the balanced equation shows that one mole of NaOH reacts with one mole of H+, the number of moles of H+ is equal to the number of moles of NaOH used:
2.430 mmol NaOH = 2.430 mmol H+
3. Determine the concentration of H+ in the 60 mL solution:
2.430 mmol H+/60 mL = 0.040 M H+
4. Calculate the total concentration of H+ in a 10 mL aliquot of Kool-Aid:
(0.040 M H+ x 0.06 L) / 0.01 L = 0.24 M H+
5. For a standard 250 mL serving of Kool-Aid:
(0.040 M H+ x 0.06 L) / 2 L = 1.2 x 10^-3 M H+
It is important to note that these calculations assume that all the acid in Kool-Aid is neutralized by NaOH and that the acid in Kool-Aid is solely represented by H+. In reality, Kool-Aid may contain other acids that contribute to its sour taste, and the pH may be affected by other factors as well.