3. in a typical titration experiment a student titrates a 5.00 ml sample of formic acid (hcooh), a monoprotic organic acid, with 26.59 ml of 0.1088 m naoh. at this point the indicator turns pink. calculate the # of moles of base added and the concentration of formic acid in the original sample.

2.893 x 10⁻³ mol NaOH

[HCOOH] = 0.5786 mol/L

Explanation:

The balanced reaction equation is:

HCOOH + NaOH ⇒ NaHCOO + H₂O

At the endpoint in the titration, the amount of base added is just enough to react with all the formic acid present. So first we will calculate the moles of base added and use the molar ratio from the reaction equation to find the moles of formic acid that must have been present. Then we can find the concentration of formic acid.

The moles of base added is calculated as follows:

n = CV = (0.1088 mol/L)(26.59 mL) = 2.892992 mmol NaOH

Extra significant figures are kept to avoid round-off errors.

Now we relate the amount of NaOH to the amount of HCOOH through the molar ratio of 1:1.

(2.892992 mmol NaOH)(1 HCOOH/1 NaOH) = 2.892992 mmol HCOOH

The concentration of HCOOH to the correct number of significant figures is then calculated as follows:

C = n/V = (2.892992 mmol) / (5.00 mL) = 0.5786 mol/L

The question also asks to calculate the moles of base, so we convert millimoles to moles:

(2.892992 mmol NaOH)(1 mol/1000 mmol) = 2.893 x 10⁻³ mol NaOH

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.