Question

Consider the reaction of 64.7 ml of 0.310 m NaC₇H₅O₂ with 50.0 ml of 0.245 m HBr. (ka of HC₇H₅O₂ = 6.3 x 10⁻⁵)

What quantity in moles of C₇H₅O₂⁻ would be present before the reaction takes place?

Answer 1

To find the moles of C₇H₅O₂⁻ present before the reaction, the volume of NaC₇H₅O₂ solution is converted to liters and multiplied by its molarity, resulting in 0.02005 moles of C₇H₅O₂⁻.

Step-by-step explanation:

We are asked to find the quantity of moles of C₇H₅O₂⁻ before the reaction takes place, with a given volume and molarity of NaC₇H₅O₂. To determine the moles, we use the molarity formula (Moles = Molarity × Volume), using the volume in liters.

First, convert the volume of NaC₇H₅O₂ from mL to liters:

• 64.7 mL × (1 L/1000 mL) = 0.0647 L

Next, calculate the moles of NaC₇H₅O₂ (which is equal to the moles of C₇H₅O₂⁻ since there is a one-to-one relationship in the dissociation of sodium benzoate in water):

• Moles of C₇H₅O₂⁻ = 0.310 M × 0.0647 L = 0.02005 moles

Therefore, 0.02005 moles of C₇H₅O₂⁻ would be present before the reaction takes place.