Question

how many milliliters of 0.15 M HCl are needed to titrate to the end point 100.0 mL of a solution that contains 2.35g of NaOH per liter?

Answer 1

Balanced chemical equation is:

NaOH + HCl ---> NaCl + H2O

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(NaOH)= 2.35 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(2.35 g)/(40 g/mol)

= 5.875*10^-2 mol

According to balanced equation

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

= (1/1)*5.875*10^-2

= 5.875*10^-2 mol

This is number of moles of HCl

use:

M = number of mol / volume in L

0.15 = 5.875*10^-2/ volume in L

volume = 0.392 L

volume = 392 mL

Answer: 392 mL