1) a student dissolved 0.1916 g of an unknown diprotic acid in 100 ml of distilled water. the acid was then titrated with 0.1025m naoh solution. the second equivalence point showed the sharpest change in ph, and so it was used to determine the molar mass of the unknown acid. the volume of naoh needed to reach the equivalence point was 27.5 ml.

a. calculate the number of moles of naoh used in the titration to reach the second equivalence point.

b. calculate the number of moles of diprotic acid, based on the fact that we are examining the second equivalence point.

c. calculate the molar mass of the diprotic acid.

a.

b.

c.

Explanation:

a. To calculate the number of moles of NaOH we need the following definition of molarity:

b. As we have a diprotic acid, it means that when the second equivalence point is reached, two moles of NaOH have reacted by one mol of acid, therefore the relation we use to determine the moles of acid is:

c. Finally, we determine the molar mass as follows:

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol