Suppose that you wanted to prepare a acetate ion/acetic acid buffer solution with a pH of 4.35. What is the value 34 Marks: 2 off [A l/IHA) for the correct buffer solution for this event? The Ka for acetic acid is 1.8 x 10 5. a. 0.39
b. 0.407
c. 0.372
d. 4.74

b. 0.407

Explanation:

Acetic acid / Acetate ion is a buffer (Mixture of a weak acid, acetic acid, with its conjugate base, acetate ion) with pKa = -log Ka = 4.74.

The simplest way to determine the pH of a buffer is using Henderson-Hasselbalch formula:

pH = pKa + log [Conjugate base] / [Weak acid]

For acetic buffer with pH = 4.35:

4.35 = 4.74 + log [A⁻] / [HA]

-0.39 = log [A⁻] / [HA]

0.407 = [A⁻] / [HA]

Thus, right option is:

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