A student was given 100mL of HCL solution in a 250mL beaker and told that the HCL was 0.11M. The student was also given phenolphthalein indicator and 250 ML of 0.1234 M NaOH. The student first filled a buret with NaOH solution. The student then placed an appropriate volume of unknown HCL into a 250 mL flask and added drops of indicator. Calculate how many of NaOH is required to complete the titration.
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To calculate how much NaOH is required to complete the titration, we need to determine the volume of HCl used in the titration. The molarity (M) of the HCl solution is given as 0.11M.
1. Start by calculating the number of moles (n) of HCl using the formula:
n = M x V
where M is the molarity and V is the volume in liters.
We have 0.11M as the molarity and the volume given is 100mL, which is equal to 0.1L (since 1L = 1000mL).
n = 0.11M x 0.1L
n = 0.011 moles of HCl
2. Now, we can determine the number of moles of NaOH required to neutralize the HCl. The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH -> NaCl + H2O
The stoichiometry of the balanced equation tells us that 1 mole of HCl reacts with 1 mole of NaOH.
Therefore, the number of moles of NaOH required will also be 0.011 moles.
3. Next, we can find the volume of NaOH required using its molarity. The molarity of NaOH is given as 0.1234M, and we already know the number of moles needed.
Rearrange the formula n = M x V to calculate the volume (V):
V = n / M
V = 0.011 moles / 0.1234M
V ≈ 0.089 L or 89 mL
So, approximately 89 mL of the NaOH solution is required to complete the titration.