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.
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!
3 answers
Bobby, cmon man
0.625 M is the answer