Question

DATA - PART 1 Write the balanced chemical equation for the reaction of HC with NaOH in the space below. NaOH + HCI - NACIT H2O Trial 4 In the table below, and asterisk (*) means e calculated quantity Trial Trial 2 Trial 3 Volume of HCl(aq) used (mL) (Diluted Toilet 10.00i 10.00ml 10.0om! Bowl Cleaner) Burette Reading: Initial Volume of 0.00ml 6.74ml 0.16ml NaOH (mL) Burette Reading: Final Volume of 26.78m1 33.75m) 27.22m! NaOH (mL) *Volume NaOH 26.78m1 27.01m1 27.06ml consumed (mL) *Moles of NaOH consumed (mol NaOH) *Moles HCI reacted (mol HCI) *Molarity of the diluted HCI solution (mol HCIL) *Average molarity of the diluted HCI solution *Molarity of the original HCI solution Show the calculations on the next page as required.

Hello there! I'm currently stuck on this next part od my lab. I'm struggling with the equations. Pleaae help! Thank you.

this is the concentrate of NaOH

Answer 1

We have the equation M1V1 = M2V2

Where M1 is the molarity of HCl, V1 is the volume of HCl

Where M2 is the molarity of NaOH, V2 is the volume of NaOH

Trail 1

10 ml $\times$ M1 = 26.78 ml $\times$ 0.1M

M1 =  2.78 $\times$ 0.1/10

= 0.2678 M

Average molarity = (molarity of trial 1+ molarity of trial 2 + molarity of trail 3)/3

To find out molarity of original solution we must know the how much we take from the solution and how much diluted

and use the equation

how much pipetted from original solution (ml) $\times$ molarity of original soution = total voume including diluted amount  $\times$ molarity we obtained