titration question:
How many moles of HCl must have been present in the 25mL of HCl solution in the 2 trials?
Given:
Trial 1: V of HCl= 25.00mL, V of NaOH used=29.50mL and M of NaOH=0.18M
Trial 2: V of HCl=25.00mL, V of NaOH used=28.50mL and M of NaOH=0.18M
Do it this way.
mols NaOH = M x L = ?
mols HCl = mols NaOH.
To determine the number of moles of HCl present in the 25 mL of HCl solution in both trials, you can use the concept of stoichiometry and the balanced chemical equation of the reaction between HCl and NaOH.
The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH -> NaCl + H2O
From the equation, we can see that the stoichiometric ratio between HCl and NaOH is 1:1. This means that for every one mole of HCl, we need one mole of NaOH to completely react.
First, let's calculate the number of moles of NaOH that were used in each trial:
In Trial 1:
V of NaOH used = 29.50 mL = 0.02950 L (since 1 mL = 0.001 L)
Molarity (M) of NaOH = 0.18 M
Number of moles of NaOH used in trial 1 = V of NaOH used * M of NaOH = 0.02950 L * 0.18 M = 0.00531 moles of NaOH
In Trial 2:
V of NaOH used = 28.50 mL = 0.02850 L
Molarity (M) of NaOH = 0.18 M
Number of moles of NaOH used in trial 2 = V of NaOH used * M of NaOH = 0.02850 L * 0.18 M = 0.00513 moles of NaOH
Since the stoichiometric ratio between HCl and NaOH is 1:1, the number of moles of HCl would be the same as the number of moles of NaOH.
Therefore, in both trials, there were approximately 0.00531 moles of HCl (trial 1) and 0.00513 moles of HCl (trial 2) present in the 25 mL of HCl solution.