A student is making a wet-mount preparation of onion epidermis cells for observation with a compound light microscope. The student cuts off a slice of onion,places it on a side, adds iodine,puts the slide on the stage, and rotates the nosepiece on the microscope to the highlife objective. using one or not complete sentence, state one error in the procedures followed by the student.

pH = 8.59

Explanation:

1,24g of benzoic acid are:

1,24g × (1mol / 122,12g) = 0,0102 moles

Benzoic acid in water is in equilibrium with benzoate ion, thus:

Benzoic acid ⇄ Benzoate⁻ + H⁺. K =  [Benzoate⁻] [H⁺] / [Benzoic acid] = 6,3x10⁻⁵

Also, benzoic acid reacts with NaOH producing benzoate ion and water:

Benzoic acid + NaOH → Benzoate⁻ + H₂O + Na⁺

The volume spent during titration is:

0,0102mol × (1L / 0,180mol) = 0,0567L. That means total volume is 0,0567L + 0,0500L = 0,1067L.

The equivalence point is reached when all benzoic acid reacts with NaOH and there is just benzoate. That is molarity of benzoate is 0,0102mol / 0,1067L = 0,0956M

The equilibrium of benzoate ion is:

Benzoate⁻ + H₂O ⇄ Benzoic acid + OH⁻ kb = kw/ka = 1x10⁻¹⁴/6,3x10⁻⁵ = 1,59x10⁻¹⁰ = [Benzoic acid] [OH⁻] / [Benzoate⁻]

The equilibrium concentrations are:

[Benzoate⁻] = 0,0956M - x

[Benzoic acid] = x

[OH⁻] = x

Replacing:

1,59x10⁻¹⁰ = x² / 0,0956M - x

1.52x10⁻¹¹ - 1.59x10⁻¹⁰x - x² = 0

Solving for x:

X = -3.8988×10⁻⁶ Wrong answer, there is no negative concentrations-

X = 3.89864×10⁻⁶

Thus:

[OH⁻] = 3.89864×10⁻⁶

pH = 14 - pOH and pOH = -log [OH⁻]

pOH = -log 3.89864×10⁻⁶ = 5,41

pH = 14 - 5,41 = 8,59

pH = 8.59

I hope it helps!