Question

what mass of NaOH is required to react exactly with 25.0 ml of 1.2 M of a triprotic acid?

Answer 1

Balanced chemical equation is:

H3A + 3 NaOH ---> Na3A + 3 H2O

lets calculate the mol of H3A

volume , V = 25.0 mL

= 2.5*10^-2 L

number of mol,

n = Molarity * Volume

= 1.2*0.025

= 0.03 mol

According to balanced equation

mol of NaOH reacted = (3/1)* moles of H3A

= (3/1)*0.03

= 0.09 mol

This is number of moles of NaOH

Molar mass of NaOH,

MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass of NaOH,

m = number of mol * molar mass

= 0.09 mol * 39.998 g/mol

= 3.60 g

Answer: 3.60 g