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aleksandrvk [35]
3 years ago
9

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 L

iOH 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 ?
Chemistry
1 answer:
Dennis_Churaev [7]3 years ago
5 0
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 - \frac{0.140 mol/L*8.55mL}{1000mL} = 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 - \frac{0.001197*1000mL}{12.35mL}
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 - \frac{0.140mol*12.35 mL}{1000mL}
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 - \frac{0.001729mol*1000mL}{8.55 mL}
molarity of LiOH - 0.202 M
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