Question

Calculate Equivalence Weight (Neutralization Equivalent) for 200.0 mg of carboxylic acid that was neutralized by 26.0 mL of 0.0945 M NaOH.

Answer 1

Balanced chemical equation is:

NaOH + CH3COOH ---> CH3COONa + H2O

lets calculate the mol of NaOH

volume , V = 26 mL

= 2.6*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 9.45*10^-2*2.6*10^-2

= 2.457*10^-3 mol

According to balanced equation

mol of CH3COOH reacted = (1/1)* moles of NaOH

= (1/1)*2.457*10^-3

= 2.457*10^-3 mol

This is number of moles of CH3COOH

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 2.457*10^-3 mol * 60.05 g/mol

= 0.1475 g

= 147.5 mg

Answer:  147.5 mg