Question

if 25.00 ml of a naoh solution of unknown concentration are titrated to a neutral endpoint with 14.11 ml of 0.500 m hcl solution, what is the concentration of base?

Answer 1

The concentration of the NaOH solution is calculated using the stoichiometry of the neutralization reaction, which gives a 1:1 mole ratio between NaOH and HCl. Given the volume and molarity of HCl, the moles of HCl and hence NaOH are determined, leading to a calculated NaOH concentration of 0.2822 M.

Step-by-step explanation:

To find the concentration of the NaOH solution, we utilize the stoichiometry of the neutralization reaction between NaOH and HCl, which is:

HCl + NaOH → NaCl + H2O

Since HCl and NaOH react in a 1:1 mole ratio, the moles of HCl will equal the moles of NaOH at the neutral endpoint. To calculate the moles of HCl, we use the given volume and concentration:

Moles of HCl = Volume of HCl (L) × Concentration of HCl (M)

Moles of HCl = 0.01411 L × 0.500 M

= 0.007055 mol

Since the moles of NaOH are equal to the moles of HCl, we can use this to find the concentration of NaOH:

Concentration of NaOH = Moles of NaOH / Volume of NaOH (L)

Concentration of NaOH = 0.007055 mol / 0.02500 L

= 0.2822 M

Thus, the concentration of the NaOH solution is 0.2822 M.