Answer:
6.25 μg/mL
Explanation:
When a dilution is made, the mass of the solute is conserved (Lavoiser's law), so the mass pipetted will be the mass in the assay. The mass is the concentration (C) multiplied by the volume (V). If the pipet solution is called 1, and the assay 2:
m1 = m2
C1*V1 = C2*V2
C1 = 250 μg/mL
V1 = 25 μL
V2 = 975 μL + 25 μL = 1000 μL (is the final volume of the assay after the addition of LDH)
250*25 = C2*1000
C2 = 6.25 μg/mL
The volume (in mL) of 0.242 M NaOH solution needed for the titration reaction is 39.44 mL
<h3>Balanced equation </h3>
CH₃CH₂COOH + NaOH —> CH₃CH₂COONa + H₂O
From the balanced equation above,
- The mole ratio of the acid, CH₃CH₂COOH (nA) = 1
- The mole ratio of the base, NaOH (nB) = 1
<h3>How to determine the volume of NaOH</h3>
- Volume of acid, CH₃CH₂COOH (Va) = 46.79 mL
- Molarity of acid, CH₃CH₂COOH (Ma) = 0.204 M
- Molarity of base, NaOH (Mb) = 0.242 M
- Volume of base, KOH (Vb) =?
MaVa / MbVb = nA / nB
(0.204 × 46.79) / (0.242 × Vb) = 1
Cross multiply
0.242 × Vb = 0.204 × 46.79
Divide both side by 0.242
Vb = (0.204 × 46.79) / 0.242
Vb = 39.44 mL
Thus, the volume of NaOH needed for the reaction is 39.44 mL
Learn more about titration:
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hope this is adequate.
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