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How can one control the enzyme experiment?

Biology

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Answer

0.1085 mol/L

Step-by-step solution

4 steps

  1. Calculate the average volume of NaOH used in the two titrations.

    Vmeˊdio NaOH=V1+V22V_\mathrm{médio\ NaOH} = \frac{V_1 + V_2}{2} Vmeˊdio NaOH=24.78 mL+25.80 mL2=50.58 mL2=25.29 mLV_\mathrm{médio\ NaOH} = \frac{24.78 \, \mathrm{mL} + 25.80 \, \mathrm{mL}}{2} = \frac{50.58 \, \mathrm{mL}}{2} = 25.29 \, \mathrm{mL}

  2. Convert volumes to liters.

    VNaOH=25.29 mL=25.29×10−3 L=0.02529 LV_\mathrm{NaOH} = 25.29 \, \mathrm{mL} = 25.29 \times 10^{-3} \, \mathrm{L} = 0.02529 \, \mathrm{L} VHCl=25 mL=25×10−3 L=0.025 LV_\mathrm{HCl} = 25 \, \mathrm{mL} = 25 \times 10^{-3} \, \mathrm{L} = 0.025 \, \mathrm{L}

  3. Calculate the number of moles of NaOH used at the equivalence point (1:1 stoichiometry with HCl).

    nNaOH=cNaOH×VNaOHn_\mathrm{NaOH} = c_\mathrm{NaOH} \times V_\mathrm{NaOH} nNaOH=0.1073 mol/L×0.02529 Ln_\mathrm{NaOH} = 0.1073 \, \mathrm{mol/L} \times 0.02529 \, \mathrm{L}

    First,

    0.1073×0.02529=0.1073×(0.025+0.00029)=(0.1073×0.025)+(0.1073×0.00029)0.1073 \times 0.02529 = 0.1073 \times (0.025 + 0.00029) = (0.1073 \times 0.025) + (0.1073 \times 0.00029) 0.1073×0.025=0.00268250.1073 \times 0.025 = 0.0026825 0.1073×0.00029=0.0000311170.1073 \times 0.00029 = 0.000031117 nNaOH=0.0026825+0.000031117=0.002713617 moln_\mathrm{NaOH} = 0.0026825 + 0.000031117 = 0.002713617 \, \mathrm{mol}

  4. Since nHCl=nNaOHn_\mathrm{HCl} = n_\mathrm{NaOH},

    cHCl=nHClVHCl=nNaOHVHClc_\mathrm{HCl} = \frac{n_\mathrm{HCl}}{V_\mathrm{HCl}} = \frac{n_\mathrm{NaOH}}{V_\mathrm{HCl}} cHCl=0.002713617 mol0.025 Lc_\mathrm{HCl} = \frac{0.002713617 \, \mathrm{mol}}{0.025 \, \mathrm{L}} cHCl=0.0027136170.025=0.10854468 mol/Lc_\mathrm{HCl} = \frac{0.002713617}{0.025} = 0.10854468 \, \mathrm{mol/L}

    Considering significant figures (4 significant figures),

    cHCl=0.1085 mol/Lc_\mathrm{HCl} = 0.1085 \, \mathrm{mol/L}

    0.1085 mol/L

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