Question

Hi ! I really need help with this calculation and its from my lab :)

Volume of HCl 20 mL
Initial volume of NaOH in buret 50 mL
Final volume of NaOH in buret when end point was reached 30 mL
Initial pH of HCl 0.70
pH after 5 mL NaOH added 0.96
pH after 10 mL NaOH added 1.3
pH after 15 mL NaOH added 2.15
pH after 20 mL NaOH added 7.00
Volume at which pH=7 20 mL
pH after 25 mL NaOH added 12.7
pH after 30 mL NaOH added 12.85
pH after 35 mL NaOH added 12.94
pH after 40 mL NaOH added 13
pH after 45 mL NaOH added 13.05
pH after 50 mL NaOH added 13.08

Answer 1

those above is my data
Determine the molarity of the HCl titrated in the demo. Include all calculations for each step of the part II lab, showing your work neatly, labeling each calculation with a subtitle, and giving a brief written explanation of each calculation.

Answer 2

So from this data you only need to determine the Molarity of the acid?

Answer 3

We also need the equation for the reaction of HCl and NaOH (neutralization), have you written it?

Answer 4

@aaronq i reallly dont know how too

Answer 5

No problem. It's simple. Any acid-base reaction makes water and a salt (meaning an ionic compound).

\(\sf NaOH+HCl→H_2O ~+ ~?\)

Can you try to guess what the salt is?

Answer 6

oKay is it NaCl ? @aaronq

Answer 7

yes it is! So we have: \(\sf NaOH+HCl→H2O + NaCl\)

Can you check if this is balanced?

Answer 8

well by the look of it i would say it is balanced by the coefficients

Answer 9

@aaronq

Answer 10

yep it's balanced as it is. Now we need to find the moles of base we used. Have you worked with molarity before?

Answer 11

yeah, right now i think the next step is to set up the stoichiometry problem so im thinking i have to convert NaOH from volumes to moles but after that im kinda stuck @aaronq

Answer 12

Okay, to find the moles of base used we use: \(\sf Molarity=\dfrac{n_{solute}}{L_{solution}}\)

I dont see the Molarity of NaOH in what you posted, though.

Answer 13

for molarity of NaOH i got 39.997

Answer 14

@aaronq

Answer 15

That is the molar mass (values you get from the periodic table) this is Molarity, a value of concentration. This mustve been told to you to perform the calculations.

Answer 16

If you dont have that value, we can shortcut all of this by using the initial pH of the acid. But that's not what the lab wants you to do, and i can't promise you you will get full marks.

Answer 17

i was actually looking thru my notes and cant find what you have asked for, so i might as well use the pH :(

Answer 18

Can you ask your lab partner or teacher?

Using the pH.

Since we know that the dissociation equation of the acid is: \(\sf HCl\rightarrow H^++Cl^-\)
and because it a strong acid, it dissociates (separates) almost 100% percent.

This means that the initial concentraition of HCl is equal to the concentration of \(H^+\) after it sits in water for a while.

The pH is given by: \(\sf pH=-log[H^+]\)

So, \(\sf [HCl]=[H^+]=10^{-pH}\)

Answer 19

By the way, the brackets \(\sf [~]\) mean Molarity.

Answer 20

okay thank you im going to wait and ask my teacher tomorrow since its online class but overall thank you for you time. @aaronq

Answer 21

thats a good idea. No problem!