what is the concentration of 42.45 ml of hcl when titrated with 25ml of 0.1 moldm3 of naoh??

Question

aaronq · Answer

at the equivalence point the moles of base are equal to the moles of acid, right?

$Molarity=\dfrac{n_{solute}}{L_{solution}}\rightarrow n_{solute}=M*L_{solution}$ 

         $n_{acid}=n_{base}$ 

so, by substitution: 

$M_{acid}*L_{solution}=M_{base}*L_{solution}$ 

which is the $M_1V_1=M_2V_2$ equation.

anonymous · Answer

what exactly does the n and l stand for??

aaronq · Answer

n is moles and L is liters

anonymous · Answer

can you also please show me how to get the no. of moles of NaOH?

aaronq · Answer

use the molarity formula (above), plug in your values for NaOH, solve for n.

anonymous · Answer

would it be 2.5 moles? sorry to ask so many questions

aaronq · Answer

it would be  n=M*L=(0.1 mol/L)*(0.025 L)=0.0025 moles

aaronq · Answer

remember that the volume is in terms of Liters, so you will need to convert.

anonymous · Answer

so that means that the no. of moles of hcl is also 0.0025?

aaronq · Answer

indeed

anonymous · Answer

is the concentration 0.0589?

aaronq · Answer

thats right. don't forget the units though

anonymous · Answer

$$moldm ^{3}$$

aaronq · Answer

it should be "mol/$dm^3$", because it's moles per liter.

anonymous · Answer

thank you so much :)

aaronq · Answer

no problem!