Suppose you performed a titration of a weak acid and you found that the equivalence point occurred at 19.19 mL of added NaOH. At what volume would you use the pH to determine the pKa of the acid?

Volume at half equivalence point, that is 19.19 mL/2 = 9.595 mL.

Explanation:

From the question we are asked to find the volume that would one can use the pH to determine the pKa of the acid and the answer will be the volume at half equivalence point.

At half equivalence point, the weak acid and the conjugate base will have the same number of moles because the number of moles of sodium Hydroxide, NaOH will neutralize half of the number of moles of the weak acid which in turn will produce more conjugate base. And this concept is what is known as the buffer solution.

HA <> H^+ + A^-

(Note: the reaction above is a reversible reaction. Also, the concentration of HA is equal to the concentration of A^-).

Therefore, we can calculate our pka from the equation below(assuming the pH is given).

pH= pka + log ( [A^-] / log [HA].

===> At half equivalence point pH= pKa.

Recall that, pka = - log ka.

Then, ka = 10^-pka.

Where pH= pKa.

Therefore, ka = 10^-pH.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

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