Question

What volume of 0.150 M HNO₃ will react completely with 70.0 mL of 0.200 M NaOH?

a) 46.67 mL
b) 70.0 mL
c) 93.33 mL
d) 140.0 mL

Answer 1

The volume of 0.150 M HNO₃ needed to react with 70.0 mL of 0.200 M NaOH is 93.33 mL, based on the stoichiometry of their neutralization reaction, which has a 1:1 mole ratio.

The answer is therefore (c) 93.33 mL.

Step-by-step explanation:

To find the volume of HNO₃ needed to react completely with 0.200 M NaOH, we can use a stoichiometry problem based on the neutralization reaction between HNO₃ and NaOH:

HNO₃ + NaOH → NaNO₃ + H₂O

This reaction is a 1:1 mole ratio between HNO₃ and NaOH. To determine the required volume of HNO₃, follow these steps:

1. Find moles of NaOH: (0.200 M) × (0.0700 L) = 0.014 moles.
2. Since the mole ratio is 1:1, moles of HNO₃ required = moles of NaOH = 0.014 moles.
3. Calculate volume of HNO₃: 0.014 moles / (0.150 M) = 0.0933 L or 93.33 mL.

The answer is therefore (c) 93.33 mL.