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What volume (mL) of 7.48x102 M perchloric acid can be neutralized with 115 mL of 0.244 M
sodium hydroxide?
O a. 375
O b. 8.60
OC. 750
O d.
125
O
188

Answer 1

The volume of 7.48x10^2 M perchloric acid that can be neutralized with 115 mL of 0.244 M sodium hydroxide is approximately 3.75 mL.

Step-by-step explanation:

The equation for the reaction between perchloric acid (HClO4) and sodium hydroxide (NaOH) is:

HClO4 + NaOH → NaClO4 + H2O

To determine the volume of 7.48x102 M perchloric acid that can be neutralized with 115 mL of 0.244 M sodium hydroxide, we can use the stoichiometry of the balanced equation.

From the equation, we can see that the ratio of moles of HClO4 to moles of NaOH is 1:1. Therefore, we can set up the following equality:

(Volume of HClO4 in mL) × (Molarity of HClO4) = (Volume of NaOH in mL) × (Molarity of NaOH)

Substituting the given values into the equation:

(Volume of HClO4) × (7.48x102 M) = (115 mL) × (0.244 M)

Solving for the volume of HClO4:

Volume of HClO4 = (115 mL) × (0.244 M) / (7.48x102 M) ≈ 3.75 mL

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