Q3) 1.5 mL from the 1.00 M HCl stock solution
Q4) 0.15 mL from the 1.00 × 10⁻¹ M NaOH stock solution
Explanation:
Q3) Determine the amount of 1.00 M HCl stock solution needed to form 100 mL of 1.50 × 10⁻² M HCl solution.
Q4) Determine the amount of 1.00 × 10⁻¹ M stock NaOH solution needed to form 100 mL of 1.50 × 10⁻² M NaOH solution.
To solve both of the questions we use the following formula:
initial concentration × initial volume = final concentration × final volume
For Q3 we have:
initial concentration = 1.00 M HCl
initial volume = ?
final concentration = 1.50 × 10⁻² M HCl
final volume = 100 mL
1.00 × initial volume = 1.50 × 10⁻² × 100
initial volume = ( 1.50 × 10⁻² × 100) / 1.00
initial volume = 1.5 mL from the 1.00 M HCl stock solution
For Q4 we have:
initial concentration = 1.00 × 10⁻¹ M NaOH
initial volume = ?
final concentration = 1.50 × 10⁻² M NaOH
final volume = 100 mL
1.00 × 10⁻¹ × initial volume = 1.50 × 10⁻² × 100
initial volume = ( 1.50 × 10⁻² × 100) / 1.00 × 10⁻¹
initial volume = 0.15 mL from the 1.00 × 10⁻¹ M NaOH stock solution
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determining molar concentrations
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