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The initial volume of HCl was 1.25 ml and LiOH was 2.65 ml. The final volume of HCL was 13.60 ml and LiOH was 11.20 ml. If the LiOH was .140 M what was the molarity of HCl ?

If the same volumes were used from question 4, but the HCl was .140 M , what would the molarity of LiOH be ?

Respuesta :

Q1)
This is a strong acid- strong base base reaction, balanced equation for the reaction is as follows;
LiOH + HCl ---> LiCl + H₂O
stoichiometry of acid to base is 1:1
volume of HCl used up - 13.60 - 1.25 = 12.35 mL
volume of LiOH used up - 11.20 - 2.65 = 8.55 mL
molarity of LiOH - 0.140 M
The number of LiOH moles reacted - [tex] \frac{0.140 mol/L*8.55mL}{1000mL} [/tex] = 0.001197 mol
according to stoichiometry, number of LiOH moles = number of HCl moles
Therefore number of HCl moles reacted - 0.001197 mol
The number of  HCl moles in 12.35 mL - 0.001197 mol 
Then number of HCl moles in 1000 mL - [tex] \frac{0.001197*1000mL}{12.35mL} [/tex]
Molarity of HCl - 0.0969 M

Q2)
Volume of HCl used - 12.35 mL
Volume of LiOH used - 8.55 mL
Molarity of HCl - 0.140 M
In 1 L solution of HCl there are 0.140 mol of HCl
Therefore number of HCl moles in 12.35 mL - [tex] \frac{0.140mol*12.35 mL}{1000mL} [/tex]
Number of HCl moles reacted - 0.001729 mol 
since molar ratio of acid to base is 1:1
the number of LiOH moles that reacted - 0.001729 mol
Therefore number of moles in 8.55 mL - 0.001729
Then number of LiOH moles in 1000 mL - [tex] \frac{0.001729mol*1000mL}{8.55 mL} [/tex]
molarity of LiOH - 0.202 M

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