Question

You are a scientist at a reputable lab researching how different polyprotic acids affect the pH of a solution. How many equivalence points would be visible on the titration curve if you fully titrated a given a solution of 1.0 M sulfurous acid (H2SO3)?

A buffer is made by adding 0.600 mol HC2H3O2 and 0.600 mol NaC2H3O2 to enough water to make 4L of solution. The pKa of the buffer is 4.74. Calculate the pH of solution after 0.035 mol of NaOH is added. (assume the volume doesn

Answer 1

a) You will see two equivalence point (one for each hydrogen)

b)

n = 0.6 HA

n = 0.6 NaA

V = 4 L

pKa = 4.74

pH after 0.035 mol of NaOH is added

The next is going to happen:

NaOH => Na+ and OH-

NaC2H3O2 => Na+ C2H3O2-

HC2H3O2 <=> H+ C2H3O2- (this is an equilibrium)

Species in solution are going to react next:

H+ and OH => H2O

Lets do the balance of how many moles of water are produced

0.6 mol of Acid and 0.035 mol of base (1:1)

0.6 mol of [H+] - 0.035 mol of [OH-] = 0.565 mol of acid left... al base is over

Since it is a buffer... we need to use henderson equation

pH = pKa + log ([Salt] / [Acid])

[Salt] = moles of salt / volume

[Acid] = moles of acid / volume

Calcualte concentrations

[Salt] = 0.6 mol of Salt / 4L = 0.15

[Acid] = 0.565 mol of Acid / 4 L = 0.1413

Substitute in Henderson Equation

pH = 4.74 + log ([0.15] / [0.1413])

pH = 4.74+ 0.026 = 4.766

Which means our buffer works (no extremely change in pH due to addition of NaOH)