The stoichiometric point of a titration is reached when 35.50 mL 0.40 M HBr is added to a 25.00 mL sample of LiOH. The original [LiOH] is 0.57 M 0.014 M 0.28 M 0.024 M
The Correct Answer and Explanation is:
To solve this titration problem, we first need to understand the neutralization reaction between HBr (a strong acid) and LiOH (a strong base):HBr+LiOH→LiBr+H2O\text{HBr} + \text{LiOH} \rightarrow \text{LiBr} + \text{H}_2\text{O}HBr+LiOH→LiBr+H2O
The stoichiometry of this reaction is 1:1, meaning one mole of HBr neutralizes one mole of LiOH.
Step 1: Determine the moles of HBr used
We are told that 35.50 mL of 0.40 M HBr is required to reach the stoichiometric point. To find the moles of HBr, we use the formula:moles of HBr=Molarity×Volume (in L)\text{moles of HBr} = \text{Molarity} \times \text{Volume (in L)}moles of HBr=Molarity×Volume (in L)moles of HBr=0.40 M×35.50 mL1000 mL/L\text{moles of HBr} = 0.40 \, \text{M} \times \frac{35.50 \, \text{mL}}{1000 \, \text{mL/L}}moles of HBr=0.40M×1000mL/L35.50mLmoles of HBr=0.40×0.0355 L=0.0142 moles\text{moles of HBr} = 0.40 \times 0.0355 \, \text{L} = 0.0142 \, \text{moles}moles of HBr=0.40×0.0355L=0.0142moles
Step 2: Determine the moles of LiOH
At the stoichiometric point, the moles of HBr added are equal to the moles of LiOH in the sample. Therefore, moles of LiOH = moles of HBr = 0.0142 moles.
Step 3: Calculate the concentration of LiOH
We are given that the volume of the LiOH sample is 25.00 mL, or 0.02500 L. To find the concentration of LiOH, we use the formula:Molarity of LiOH=moles of LiOHvolume of LiOH (in L)\text{Molarity of LiOH} = \frac{\text{moles of LiOH}}{\text{volume of LiOH (in L)}}Molarity of LiOH=volume of LiOH (in L)moles of LiOHMolarity of LiOH=0.0142 moles0.02500 L=0.568 M\text{Molarity of LiOH} = \frac{0.0142 \, \text{moles}}{0.02500 \, \text{L}} = 0.568 \, \text{M}Molarity of LiOH=0.02500L0.0142moles=0.568M
Step 4: Compare the calculated value to the given options
The calculated concentration of LiOH is 0.568 M, which is closest to 0.57 M.
Thus, the correct answer is 0.57 M.
This means the original concentration of LiOH in the sample is 0.57 M. The titration process helped us determine this by using the stoichiometric relationship between the acid and base in the neutralization reaction.
