Question

3. If 45.0 mL of 0.560 M KOH are needed to neutralize 125 mL of HCl of unknown concentration, what is

Answer 1

Answer:

Molarity = 0.202 M

7.36 g/L HCl

Explanation:

KOH                        +                  HCl ----> KCl  +  H2O

1 mol                                           1 mol

0.560mol/L *0.045L                 x mol/L*0.125L

0.560mol/L *0.045L    =   x mol/L*0.125L

x = 0.560mol/L *0.045L  /0.125L= 0.2016 mol/L≈ 0.202 mol/L=0.202M HCl

M(HCl) = 1.0+35.5 = 36.5 g/mol

0.2016 mol/L*36.5 g/mol ≈ 7.36 g/L HCl