The simplest carboxylic acid is methanoic acid HCOOH. It occurs naturally in ants.

Question

The simplest carboxylic acid is methanoic acid HCOOH. It occurs naturally in ants.

When an ant bites, it injects a solution containing 50% by volume of methanoic acid. A typical ant may inject around 6.0 x 10-3 mL of this solution.

a) When you are bitten by an ant it does not inject you with all of its methanoic acid but keeps a little in reserve. Assuming that a typical ant injects 80% of its methanoic acid, what is the total volume of pure methanoic acid contained in a ‘typical ant’?

b) How many ‘typical ants’ would have to be distilled to produce 1 L of pure methanoic acid?

Sodium hydrogencarbonate, NaHCO3 is often used to treat ant stings. The reaction between methanoic acid and sodium hydrogencarbonate is:

HCOOH + NaHCO3àHCOO- + Na+ + H2O + CO2

c) Given that the density of methanoic acid is 1.2 g mL , how many moles of methanoic acid

does the typical ant inject?

d) What mass of sodium hydrogencarbonate would be required to neutralise completely the sting from this ant?

Answer 1

Hello Zeee, this is a cool problem. Let's break this up into steps:

a. 6.0 x 10^-3 ml solution is injected and is 50% by volume of HCOOH

So the ant injects HCOOH that makes up half this volume of solution which means 3.0 x 10^-3 mL of HCOOH

The ant injects only 80% of its original amount of HCOOH.

If this amount is 3.0 x 10-3, then then the original amount is

3.0 x 10^-3 x 100/80 = 3.75 x 10^-3 mL. This is the amount in a typical ant

b. To make 1 L of HCOOH, use a ratio-proportion concept: (convert L to mL first, 1 L is 1000 mL)

To make 3.75 x 10^-3 ----------> 1 ant

To make 1000 mL -----------> 1000/ 3.75 x 10^-3 = 2.6 x 10⁵ ants

HCOOH + NaHCO₃ --> HCOONa + CO₂ + H₂O

Check if equation is balanced. (It already is)

c. In order to determine moles we need mass first

density = mass/volume. So, mass = density x volume

mass of HCOOH = density x volume

= 1.2 g/mL x 3.0 x 10^-3mL = 0.0036 g or 3.6 x 10^-3 g

Next, calculate formula mass of HCOOH. 2 H, 1 C and 2 O make 46.02 g/mol

mol HCOOH = 0.0036 g

---------------

46.02 g/mol

= 7.8 x 10^-5 mol (using only 2 sig. fig)

Since HCOOH and NaHCO₃ react in a 1 : 1 ratio (the balanced equation has 1 as coefficent for both)

we need the same amount of NaHCO₃ i.e 7.8 x 10^-5 mol

To calculate mass, we need formula mass of baking soda.

1 Na, 1 H, 1 C 3 O atomic masses add to ( 22.99 + 1.0 + 12.0 + 48.0 ) = 84.0 g

Multiply mol with formula mass to get actual mass

7.8 x 10^-3 x 84.0 = 0.6652 g or 0.66 g of baking soda

Next time an ant bites, I will make sure to keep some baking soda handy!