If 25 mL of a 0.5 M NaOH solution were needed to reach the equivalence point of a 20 mL HCl solution, what was the initial analyte concentration?
I am confused how to solve this problem, and also, what does "initial analyte concentration" refer to? Thanks!
0.625 M is the answer
Bobby, cmon man
I've never seen that phrase used in a problem like this where there was no dilution but i assume they want to know the concn of the HCl.
There is a formula you can use like this.
M1V1 = M2V2
0.5M x 25 mL = M2 x 20 mL and solve for M2. That formula is good only for 1:1 reactions such as NaOH with HCl or H2SO4 with Ca(OH)2.
The following will work with any.
1. Write and balance the equation.
NaOH + HCl ==> H2O + NaCl
2. mols NaOH - M x L = ?
3. Using the coefficients in the balanced equation, convert mols NaOH to mols HCl. In this case they are equal.
4. Now, M HCl = mols HCl/L HCl.
To solve this problem, you can use the concept of stoichiometry and the balanced equation between the acids and bases involved. The initial analyte concentration refers to the concentration of the acid (in this case, HCl) before it reacts with the base (NaOH).
To find the initial analyte concentration, you need to set up an equation using the stoichiometric ratio between the acid and base. The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H2O
From the equation, you can see that the stoichiometry is 1:1, which means that one mole of HCl reacts with one mole of NaOH to produce one mole of NaCl and one mole of water.
First, determine the number of moles of NaOH used at the equivalence point. The volume of NaOH solution used is 25 mL, and the concentration is 0.5 M. Using the formula:
moles NaOH = concentration x volume
moles NaOH = 0.5 M x 0.025 L
moles NaOH = 0.0125 mol
Since the stoichiometric ratio is 1:1, the number of moles of HCl used is also 0.0125 mol.
Next, use the volume of the HCl solution to calculate its initial concentration. The volume of the HCl solution used is 20 mL, which is equivalent to 0.02 L.
initial analyte concentration = moles HCl / volume HCl
initial analyte concentration = 0.0125 mol / 0.02 L
initial analyte concentration = 0.625 M
Therefore, the initial analyte concentration of the HCl solution is 0.625 M.