The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.150 M NaOH. It requires 21.4 mL of the NaOH solution to reach the end point of the titration. A buret filled with a titrant is held above a graduated cylinder containing an analyte solution. What is the initial concentration of HCl

0.321 M

Explanation:

The neutralization reaction of HCl with NaOH is the following:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

So, 1 equivalent of HCl reacts with 1 equivalent of NaOH. In this case:

1 equivalent HCl = 1 mol HCl

1 equivalent NaOH = 1 mol NaOH

At the equivalence point, the amount of HCl reacts completely with the added amount of titrant (NaOH). As 1 mol of reactant is calculated from the product of the molarity of the solution and the volume, in the equivalent point:

moles HCl = moles NaOH

M(HCl) x V(HCl) = M(NaOH) x V(NaOH)

We have:

M(NaOH) = 0.150 M

V(NaOH) = 21.4 mL = 0.0214 L

V(HCl) = 10.0 mL = 0.01 L

Thus, we calculate the initial concentration HCl as follows:

M(HCl) = (M(NaOH) x V(NaOH))/ (V(HCl))

            = (0.150 M x 0.0214 L)/(0.01 L) = 0.321 M

the answer for sodium is 0.186

the answer for rubidium is 0.251