# 0.3 g of citric acid is dissolved in distilled water to produce 100 mL of 0.016 M of solution. How much citric acid is needed in order to complete a full neutralization process with 30 ml of 0.5 M...

0.3 g of citric acid is dissolved in distilled water to produce 100 mL of 0.016 M of solution. How much citric acid is needed in order to complete a full neutralization process with 30 ml of 0.5 M sodium bicarbonate solution? What changes would you make in order for the reaction to run faster?

pnrjulius | Certified Educator

To make this reaction run to completion, we want to ensure that there are the right number of moles of each reactant so that none will be left over.

The equation for citric acid neutralization with sodium bicarbonate is as follows:
`H_3 C_6 H_5 O_7 + 3 NaHCO_3 rightarrow 3 CO_2 + 3 H_2 O + Na_3 C_6 H_5 O_7`
That is, 1 mole of citric acid plus 3 moles of sodium bicarbonate produces the reaction. So we need 3 times as many moles of sodium bicarbonate.

We currently have 100 mL of 0.016 M citric acid, which is `(0.016 {mol}/{L})(0.1 {L}) = 0.0016 mol` , or 1.6 mmol.

We also have 30 mL of 0.5 M sodium bicarbonate, which is `(0.5 {mol}/{L})(0.03 {L}) = 0.015 mol` , or 15 mmol.

So, to match the 15 mmol of sodium bicarbonate, we need a total of 5 mmol of citric acid. That means we need an additionalÂ 3.4 mmol.

Since 0.3g of citric acid produced 1.6 mmol, that additional 3.4 mmol will require `(3.4/1.6)(0.3 g) = 0.6375 g` , or 637.5 mg.

Once we add the 637.5 mg to the 100 mL of 0.016 M solution, we will have 5 mmol in 100 mL, which is a 0.05 M solution.

Thus we mix the 100 mL of 0.05 M citric acid with the 30 mL of 0.5 M sodium bicarbonate, matching 5 mmol of citric acid to 15 mmol of sodium bicarbonate.

This is an endothermic reaction, so it will proceed faster if we heat the solution to a higher temperature (thus providing the energy needed to sustain the reaction). Physically stirring the solutions together to mix them better would also speed up the reaction somewhat.