Students performed a procedure similar to Part II of this
experiment (Analyzing Juices for Vitamin C Content) as described in the
procedure section. Given that molarity of DCP is 9.98x10-4 M, it took 16.34 ml
of DCP to titrate 10 mL of sample.
Amount of ascorbic acid = 0.050 L sample (0.01634 L DCP/0.01
L sample)( 9.98x10-4 mol DCP/L DCP)(1 mol Ascorbic acid/ 1mol DCP)(176.124
g/mol)(1000mg/1g)= 14.36 mg ascorbic acid
I believe that the answer is A i could be wrong though.
Any organism in motion possesses kinetic energy. Kinetic energy is the energy the body has while in motion.
When Ka =1.37 x 10^-4
∴Pka = - ㏒Ka
= -㏒ 1.37 x 10^-4
= 3.86
According to H- H equation:
when PH = PKA + ㏒[conjugate base / weak acid]
when the lactate is the conjugate base and lactic acid is the weak acid
∴ PH = Pka +㏒[lactate] / [lactic acid]
so, by substitution:
4.49 = 3.86 + ㏒[lactate]/[lactic acid]
∴[lactate]/[lactic acid] = 4.3
∴ [lactic acid] / [lactate] = 1/4.3
= 0.23
