Question

2. Calculate the number of grams of HCI in 100.0 mL of concentrated HCl which is approximately 12.1 M and 37 % HCl by weight and the density of 1.191 g/mL. 3. An acid-base titration is performed but some base that splashed onto the sides of the flask is not rinsed down prior to the endpoint measurement. Will the calculated molarity of the acid sample be erroneously high or low? Explain. 4 An acid solution ir titrated to thanimahallalaid in hun harcolution time

Answer 1

Molarity is the number of moles of solute present in one litre of solution.

$M = \frac{Weight of solute}{Molar mass of solute}*\frac{1000}{Volume of the solution in mL}$

Weight of solute is given by$Weight of solute = \frac{Molarity of solution* volume of the solution* molar mass of solute}{1000}$

Given that Molarity of HCl = 12.1 M

Weight of HCl = 12.1 * 100 * 36.5/1000

= 44.17 grams

2) When added base is splashed on the walls of the flask, more amount of base required to neutralize the given volume of acid. The concentration of acid is higher than the theoretical value.

M1V1/n1 = M2V2/n2

M2 = M1*V1*n2/n1*V2

Since V1 is the volume of the base and other parameters are same, concentration of resultant acid is higher than the actual value.