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?
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
1 answer