What volume of 0.812M HCl is required to titrate 1.33g of NaOH to the endpoint?

Answer 1

First, look at the balanced equation:

HCl + NaOH ==> NaCl + H2O

Next, we find the moles of NaOH being neutralized by the HCl. Then that will be equal to the moles of HCl needed since the balanced eq. shows they react in a 1:1 mole ratio. Finally, we'll convert moles of HCl to mls of HCl.

moles NaOH present = 1.33 g x 1 mol NaOH/40 g = 0.03325 moles NaOH

moles HCl needed = 0.03325 moles NaOH x 1 mol HCl/mol NaOH = 0.03325 moles HCl needed

volume HCl needed: since the HCl is 0.812 moles/L we use this to find liters or mls needed to provide 0.03325 moles.

(x L)(0.812 mol/L) = 0.03325 moles

x = 0.0409 L

x = 40.9 mls (to 3 significant figures)

Answer 2

HCL+NaOH= NaCl+H2O this reaction of neutralization

Titration is a process based on the fact that known reactant completely react with other reactant which will be gradually added to solution.

Molar mass of HCl is 36,5

Molar mass NaOH is 40

First, let's find from proportion, how many gram HCl will react with 1,33 g NaOH

(36.5x1.33) :40= 1,21 gram

But we need to find the volume of solution of HCl

1M solution of HCl equal 36.5 HCl, but 0.812 M equal 29.64 g

Now we can find out the volume of solution in liters where will be dissolved 1.21 g HCl

We found 1 L of solution 0,812 M holds 29.64 g HCl, but 1.21 gram HCl dissolved in 41 ml solution.

1.21:29.64=0.041 L= 41 ml