Question

# A NaOH titrant solution with 2 grams of NaOH in 100mL where the Volume is 40mL, was used to determine the concentration of HCl acid $analyte solution$ where the volume = 20mL. Calculate the concentration of HCl acid analyte solution in mol/L. Present your solution in mol/L to 1 significant figure. Enter a numeric value only without the mol/L unit.

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## Answer to a math question A NaOH titrant solution with 2 grams of NaOH in 100mL where the Volume is 40mL, was used to determine the concentration of HCl acid $analyte solution$ where the volume = 20mL. Calculate the concentration of HCl acid analyte solution in mol/L. Present your solution in mol/L to 1 significant figure. Enter a numeric value only without the mol/L unit.

Birdie
4.5
Given:
- Mass of NaOH = 2 grams
- Volume of NaOH solution = 100 mL = 0.1 L
- Volume of HCl solution = 20 mL = 0.02 L
- Volume of NaOH titrant solution = 40 mL = 0.04 L

1. Calculate moles of NaOH:
\text{Moles of NaOH} = \frac{2 \, \text{g}}{40 \, \text{g/mol}} = 0.05 \, \text{mol}

2. Calculate the molarity of NaOH $M_1$:
M_1 = \frac{0.05 \, \text{mol}}{0.1 \, \text{L}} = 0.5 \, \text{mol/L}

3. Use the titration formula to find the molarity of the HCl solution $M_2$:
M_1V_1 = M_2V_2
0.5 \, \text{mol/L} \times 0.04 \, \text{L} = M_2 \times 0.02 \, \text{L}
M_2 = \frac{0.5 \times 0.04}{0.02} = 1.0 \, \text{mol/L}

Therefore, the concentration of the HCl acid analyte solution is approximately \bf{1.0} mol/L $rounded to 1 significant figure$.

\boxed{1.0}

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