Question

How much 0.670 M HCl do you need to neutralize 45.8g NaOH dissolved in 1.22L of water?

Answer 1

we have the Balanced chemical equation as:

NaOH + HCl ---> NaCl + H2O

Molar mass of NaOH = 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 = 45.8 g

we have below equation to be used:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(45.8 g)/(39.998 g/mol)

= 1.145 mol

From balanced chemical reaction, we see that

when 1 mol of NaOH reacts, 1 mol of HCl reacts

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

= (1/1)*1.145

= 1.145 mol

This is number of moles of HCl

we have below equation to be used:

M = number of mol / volume in L

0.67 = 1.145/ volume in L

volume = 1.71 L

Answer: 1.71 L